I like using them in my math class to illustrate how the wrong method can still get the right answer on occasion, and show the common error of improper "cancellation" to simplify fractions.
At 6:12, the reason you get a nonsense answer here is independent of the (a^b)^c = a^(bc) step. You've stated that 1^(1/2) = 1, which is correct, used (a^b)^c = a^(bc) to find that 1^(1/2) = -1, which is also correct. The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another. If you are careful to always note that e^f(x) = e^(f(x) + 2 i n pi), where n is an integer, then that avoids the problems associated with (a^b)^c = a^(bc). Of course, using this method isn't helpful when the exponent has a base of 1, because using a base of 1 is generally problematic and breaks a lot of general rules. For example, 1^(infinity) is undefined rather than infinite.
1^(1/2) = 1 It doesn't have 2 solutions, you're mistaking it with x^2=1 which would be |x|=1^(1/2) the x is inside a module so there you have 2 solutions.
"The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another." - ya, because the sqrt function has 2 branches, one of which is the principle one
(a^b)^c is indeed not always applicable to complex numbers, however, your example is *wrong*. 1 = e^2PI*i then you raise both sides to the power of 1/2, effectively taking the square root. The square root function is not injective, you should have added +- to one of the sides, which would have made the equation correct. Basically you've shown that both 1 and -1 are square roots of 1 (which is not a mistake)
Agree, but in my opinion, (a^b)^c is applicable in case of purely 1 dimensional calculations. For example, here, We are purely using y-axis, so using this formula gives correct answer. When it comes to 2D calculations, like vectors (in complex number when both y axis and x axis are involved), this formula should become invalid, or should be generalised to get proper answer.
@@raffaeleraia2223 The thing is, square root of 1 truly have two values 1 and -1. But, we defined it as only positive 1. In the equation it will always give two values. And, a mathematical solution doesn't necessarily follow what we defined, unless we change the solution to make it follow what we defined. I am not good at words here it seems.
@@imfuture7543the reason square root usually gives a positive and not a negative is due to the fact that having just a single output is useful. so to do that the principal value is given
Why does being injective matter ? The square function is also not injective and 1 = x => 1² = x² is still true. Adding +- would be extremely wrong too since x -> x^½ is a single value function
it depends on how you define complex logarithms. if you want to have a function, you need to allow a principal value of the logarithm. if you want to find all solutions, it isn't a normal function anymore because it has multiple outputs. method 2 isn't a "wrong" proof, it just glosses over some small detail with +2*k*pi*i in the logarithm function which does cancel out when you raise it to e. It's obviously different from the fraction coincidence at the beginning because this is a good method that always works for any similar problem. you can proof euler's theorem with taylor series or you could just assume the result. doesn't make it wrong.
Actually the 2kπi won’t cancel out because you have an extra i, so raising it to e would give you a factor of e^(-2kπ), which contributes to the multi-valued nature of i^i itself.
There is nothing wrong with the first two proofs. Those are functions that give multiple branches, which is why you can produce an apparent contradiction, but in each case the calculation does still give _A_ branch, and that branch is a valid evaluation.
However, choosing a specific branch without giving a reason why, is still wrong. The first proof in fact technically falls into the fallacy mentioned with canceling digits - for example, canceling digits works as long as the number of trailing zeros in the numerator is greater than or equal to the number of trailing zeros in the denominator, and the denominator is a power of 10, and the only digit being canceled is 0. But still it's an unjustified way to prove for the others. The second one is more like you're saying, but still, choosing a branch requires justification, otherwise you can never be sure that you chose the correct branch. So it is technically wrong (but incomplete is a better description)
@@hbrg9173That's basically the principal branch, and taking the principal branch by default is reasonable to me. I would say these proofs are to introduce the concept of index form rather than slam dunking the audience terms of complex analysis which is complex as u see by its name.
@@d3ds1r There are two topics. For solving equation, it's totally fine to have multiple answers. But if you want to define a function, You usually apply a polar form and construct a Riemann Sphere to try to make an only output.
Your counterexample to proof 1 is finding out that 1 has two square roots, one of which is 1, and the other of which is -1. This is not wrong. That's always something you need to consider when undoing a root with a square! You're focusing on an extraneous root.
In general though, the core of this is that when you're using multivalued complex functions, yes, you do in fact need to be careful because the principal value isn't always the one you're looking for! This is why the proof 2 got the nonsensical result of 0 = 2pi*i So yes, the careful analysis is accurate.
@@sylverfyreThe simplest solution is to note that polar representations are not unique. The issue is not the algebraic operation in the exponent as he makes it seem. That is perfectly fine. And roots are taken care of by applying De Moivre’s formula. (Actually even complex roots are no trouble since ℂ is a field and we can always rewrite a rational function in i as a linear polynomial in i.)
with complex numbers, the log function and fraction exponents are multivalued. in the steps where you "disproved" rules of math for complex numbers you conveniently decided to ignore the multivalued nature of these operations, which you can not do.
The video is a nice explanation of branch cuts and all, but I think a lot of it can be avoided by simply noticing that a complex number has more than one polar representation. Then the first “proof” works trivially. I mean, ℂ is a characteristic 0 field for crying out loud. That’s a pretty nice algebraic structure as far as I’m concerned.
Thank you, I was getting so mad watching those "disproofs." Conveniently ignoring the multivalued nature of roots and logarithms to say things are wrong. Literally every video (and in english like he mentions) I've seen showing i^i is real either explains that they're taking principle values of everything meaning that ln(e^2ipi) would _indeed_ be 0, or they specifically mention that it's multivalued and get the final result with n somewhere indicating its multivalued nature.
I love complex analysis! In the case of the natural logarithmic function, we can make a branch cut and define the principal branch as log z = ln r + iθ, where r > 0 and θ is in (-π, π]..
Cutting a locally holomorphic function kinda defeats the pupose of it being locally holomorphic in the first place. Path dependence is just the way to go in complex analysis. If cutting is required, you would want to choose the cut freely anyway. My favourite aspect about mathematics is that any two people independently discovering the same subject should derive equivalent definitions.
I mean, the second proof was basicaly showing that e^(-pi/2) is an answer. The only thing it did differently from the final proof was considering just 1 answer to ln(x), but this answer is still valid so it should also get only one answer to i^i (and it did)
doesnt the 1=-1 proof fail because you decided that squareroot 1 = 1 instead of squareroot 1 = +-1 ? For complex numbers, if you account for all branches the identity (a^b)^c=a^(bc) should hold.
likewise in proof 2 counterexample, the natural log needs to be defined respective to a branch of the natural log which effectively reduces it mod 2πi. if you want to avoid that you need to choose a branch which includes 0 properly, so you can let πi =-πi and then you wouldnt get 0=2πi rather 0=0
@@TheEternalVortex42 That's actually not true, although Presh's example doesn't make that clear. There is an example due to Clausen where the formula fails, even when considering the multi-valued case. (See the Wikipedia page "Exponentiation" and search for "Clausen").
@@TheEternalVortex42I agree that that is the point of the video. But I disagree that that is an actual typical problem. I feel like he was assuming presentstions have the problems he's saying, but they just don't.
When, in "proof" 1, you substitute 1^(1/2) for 1, you must also say that 1^(1/2) has two results, namely 1 and - 1. So the power law holds for one of the values.
No, "log" is used for log-base-10 (though some weirdos at ISO have decided that "lg" means base-10 when that is more commonly used for base-2...the idea is that "log" means the generic logarithm where you need to provide the base, making the binary logarithm "lb.") Are you thinking of "Log", with a capital L?
@@dhwyllit depends on the field of study. In some texts that focus on graph theory and computer algorithms, they write log and it is assumed to be based 2. In some complex analysis texts, log is base e. You have to adopt the convention of the author and the field of math you are working in.
@@dpgoulette Exactly 😊 At high school, we used the names ln, log and log_a, for e-based, 10-based and a-based logarithms. At the university we used (almost) only the natural logarithm and named it log. That is: Check the definition of log in the text you are studying, and if the definition is not provided you have a natural choice 😂.
Dam this definitely blew my mind🤯 I have been so used to doing these problems normally that I almost forgot the limits of the basic mathematics rules like this indices one!
The multi valued result is still ridiculous. It means that all these values are equal to i^i, but not equal to each other. This violates the postulate that if a=b and b=c the a=c.
8:00 The natural log of any number except zero has infinite solutions. When you take the ln of 1, it would give the following solutions: ln(1) = 0 + n*2*pi*i, where n is an integer. Similar to ln(e^(2*pi*i)), it would result in: ln(e^2*pi*i) = 2*pi*i + n*2*pi*i = (n+1)*2*pi*i = m*2*pi*i, where m is an integer. This means: ln(1) = ln(e^(2*pi*i)).
@@mr.d8747the nth root FUNCTION is defined using the principal branch. Therefore, only 1 value comes from it. The 1/n exponent takes all branches. It's a multivalued function and you have to specify the branch
Than explains so much! Thank you! No matter if it's i^i or whatever else, the fact you made me pay more attention while carrying out supposedly obvious operations is priceless.
you have taken 2nd principal value for 1 which will give you -1 its no surprise there as 1^1/2 may equal to -1. but for 1=e^0 1st principal value it will remain 1. Well new method is also correct but the previous method also works we just need to be careful. as we have applied the the method to find roots of unity x^5=1.
Presh, you're usually right, but this time... All the values of the "counter-proofs" are a bit of fraudulent. In the first, you took not the principal value of one in the complex plane but another one (1=e^2πi) and you stated that is equal to one (true). But then you took the square root. 1 is the principal complex square root of the complex number 1, but you took the complex square root (which is multi-valued) of e^2πi. In other terms, you took the real square root on the LHS (which matches with the principal complex square root on complex numbers with imaginary part = 0) and the complex square root (multivalued) in RHS. If you treated 1 with the general argument (e^i(0+2πk), k is an integer) you wouldn't get to this result because it has another value which is 1. In the other, happens the same thing. You took the real natural logarithm on the LHS (which matches the principal branch of the natural logarithm on complex numbers with imaginary part = 0) and the complex natural logarithm on the RHS (the multi-valued natural logarithm). That property has to be true, not also because you use it when solving the equation z = e^w, but because is the definition of the natural logarithm (and it has to be multi-valued because so it is the complex exponential) In both cases, you took two different functions on the two sides, making the answer blatantly wrong. In complex numbers, is better to work with the general argument and then reduce it to the principal form if you're not sure what you're doing. Technical misunderstandings: Ln(z) is the notation of principal branch of natural logarithm, usually not the notation for real natural logarithm. It's subtle because for real numbers is equal, but not for complex numbers. Use Log(z) and log(z) notation for complex natural logarithm (since defining based-logarithms is a bit useless in the complex world) and ln(x) for natural real logarithm. Thank you for reading to anyone who kept reading so far and I love you Presh, this is just constructive criticism
Slight correction on proof 2: Since e^{2i\pi} = e^{0}, and in fact any angle shifted by 2\pi is the same angle on the unit circle, it can be said that 0 is in fact equal to 2i\pi. Or more precisely, this is a number system in which all numbers in the exponent are modulo 2\pi, and mod(2\pi, 2\pi) = 0. It is still interesting to see where these steps can go wrong, so thank you for the informative video.
the step at 13:25 states that the modulus of i is equal to 1 but when we will open the modulus we get : i = ± 1 that means √-1 = ± 1 which is INCORRECT!
The whole cancelling digits thing is really interesting. I want to know what process can lead you to find a general solution for what fractions that happens to
Or, you could just do it directly. The formula for complex exponentiation is: (a + bi)^(c + di) = (a^2 + b^2)^[(c + di)/2] * e^[i(c + di) * arg(a + bi)] Where arg(a + bi) is the angle with the positive real axis for the number. Thus, for i^i, that is (0 + i)^(0 + i). Therefore, a = 0, b = 1, c = 0, d = 1. Thus: (0 + 1i)^(0 + 1i) (0^2 + 1^2)^[(0 + 1i)/2] * e^[i(0 + 1i) * arg(0 + 1i)] 1^(i/2) * e^[i(i) * (π/2 + 2πn)] 1 * e^-(π/2 + 2πn) e^-(π/2 + 2πn)
Sorry, got lost in the middle of the whole thing. IMHO you also fell in the pitfall you wanted to avoid, and this started the moment you presented i in polar form. The real problem, before you get to i^i is to explain, what it means to elevate anything to a power of a complex number. What I mean is that sometimes the meaning of some mathematical operators become a little esoteric. For instance a+b is a + 1 + 1 + 1 ...+1 (b times) a * b is a + a + a ... (b times) +a. But when it get to exponentiation, and we start having a^1/2, it starts being something else than a * a (half a time?). The notation starts taking a special meaning. And this is a simple point that could be better explored in the video.
It's amusing that all the comments point out that the counterexamples are only "wrong" because they ignore that complex log is multivalued. Yet the whole point of that part of the video is that other videos are ignoring that fact!
At 6:00 it was you that messed up. 1 = e^(2*pi*i) you were fine, but when you took the square root of both sides you messed up. -1 is also equal to 1^(0.5). You played a trick with the multi-valued nature of the square root function. The issue has nothing to do with the "multiply the exponents" rule - it was another mistake altogether. To put it another way, when you introduced the 1/2 exponent, you took the principle value on the left but the other value on the right.
presh still amazes his audience with his mastery or mathematics both on the elementary and advanced levels: presh, once again, challenges me to catch-up and review mathematic principles: touché presh
I see a lot of people saying that he ignored +/-. He did not. He is using the function definition of sqrt, which is what you normally use for proofs. Using the operator definition is only fine for solving an equation (in which case you have to test for pseudo-answers). The reason the operator definition can't be used in proofs is because sqrt(1)=+1 and sqrt(1)=-1 is true in an equation, but in a proof, this would mean that per the transitive property of equality 1= -1, which is not true.
Technically speaking if we look at negative 1 and positive 1 as being unit vectors, their magnitudes are absolutely equal, it's just that their directions are point away from each other along the same line. The radian measure from 1 to 1 is + / - PI radians or + / - 180 degrees. The arccos of the dot product between the points (1,0) and (- 1,0) through the definition of the cosine function is PI radians or 180 degrees. So in some ways 1 and - 1 are kind of equal. This is why we can write them as + / - 1 which is referred to plus or minus one. The operative logical word here is OR. This is why the absolute value function always returns a positive value for all non zero numbers. If we are concerned with the actual forward direction of the vector then the above statement wouldn't be + / - 1, plus or minus one, instead it would be plus AND minus one where AND would return false. Within the first context of plus OR minus, if either is true then the entire statement is true. How is this possible? How can - 1 and + 1 simultaneously be equal to while not being equal to each other? Well think of | + / - 1| as being an arbitrary unit vector. Stand in place facing north and reach your arm out in front of you. Your looking direction is forward. The length of your arm being an arbitrary unit of measure is worth 1 unit of your arms length. Now while keeping your are stretched out in front of you turn 180 degrees or PI radians to face south. Your arm being a unit length away from your body being the center or focal point is the radius of a half circle. Your arm created an arc, a curved line. The radius of this arm, the length of your arm did not change. It is equivalent the entire time. What has changed was your facing direction. You were pointing north, and now your are point south. The points at due north and due south make either a straight line, or a curved geodesic if on a multi dimensional curved surface such as a sphere or a cylinder. This is why polar coordinates work in the way they do, this is why the complex numbers exhibits the properties of rotations and they are easily convertible. The value doesn't change, the direction does. Even within every day use such as with currency with credit and debt... If someone gives you a dollar you gained a dollar and you are +1 in the books. If you owe a dollar to someone else to where you have to give it away, you are at a loss and are - 1 in the books. The value of the dollar is still 1. The direction it is going is what changed. These directional differences between 1 and - 1 are parallel directions. They're not orthogonal. Orthogonal would be 1/2 of that rotation or 1/2 of that direction. Orthogonal directions would be East and West or Up and Down from North and South. This is why the multiplication of i is equivalent to a 90 degree or PI/2 radian rotation. Yes, we've been calling them imaginary and complex numbers for years... In truth I think this nomenclature is truly an inaccurate description. The complex terminology is just fine the way it is because the complex numbers are just that, they are complex because there is a real part and there is a secondary part that we have been calling imaginary. I think what ought to be changed, but probably won't happen because of the billions invested in writing all of the text books... would be to accurately call the imaginary numbers the orthogonal or perpendicular numbers. That's what they are. They are orthogonal - perpendicular to the real numbers. They are rotated 90 degrees or PI/2 radians from the real number plane. The square root, well in fact, the even roots of all negative numbers are orthogonal to the reals. This is why there are even and odd functions both within the polynomials as well as the trigonometric functions. x^2, x^4, x^6, are even and x, x^3, x^5, etc... are odd. The cosine and secant are even functions, and the sine, cosecant, tangent and cotangent are odd functions. Yet the sine and cosine functions have the same exact wave form, same period, same range and domain, the same properties of their limits, the only difference is their initial starting positions, The sine starts at (0,0), and the cosine starts at(0,1). The points (0,0) and (1,0) are 90 degrees from each other. This makes a vertical line which is perpendicular to the x-axis or the domain of the given function. Also their graphs are 90 degree or PI/2 horizontal translations of each other, they are 90 degrees out of phase. If we dive a bit deeper we can see that both the Pythagorean Theorem and the equation of the circle for most tense and purposes are the same thing, it's just that the Pythagorean Theorem is within the context of Right Triangles and the Equation of the Circle relates the center of it's circle, it's radius, and the arc that it generates. This is why the trigonometric functions have Pythagorean Identities. When we dive in to Physics especially within the realm of electricity or the electromagnetic spectrum, the real and "imaginary" parts or what I like to call the orthogonal parts of the waves functions is quite intriguing. This can also be seen within optics. It pertains to wave propagation. So if we now facing South and rotating in the same turning direction as before and rotate another 180 degrees or PI radians... We are now facing north again. You are pointing in the original direction, you haven't displaced yourself (horizontal linear transformation of translation), you only rotated in place, and the length of your arm hasn't changed. We have just went full circle, and this is why things such as reflections and symmetry exist. Think of + 1 and - 1 as being orthonormal reflections of each other meaning that they are pointing away from each other along the same line of sight from a common center point. If they were facing the same directions, the arccos would 0 as in 0 radians or 0 degrees (here we don't included 2PI or 360 because of the range and domain of the arccos function, and if they were pointing towards each other then the arccos would then produce? I think this would be another complex number... So that's how two values can simultaneously be equivalent yet not equivalent to each other. They have the same magnitude or absolute value but their sign, direction, or heading is different. And what's quite interesting is if you graph both the arccos and arcsine functions together it draws the outline of an hour glass, and it also takes on the representative shape of the double helix or the structure of DNA...
Saw some people disagreeing with the argument against the “proofs”, so here’s a bit of a take of mine. Regarding counterexample in “proof” 1: The exponential function is single-valued, in fact holomorphic, on all of C. Therefore all the steps in the counterexample, except where the erroneous Exponentiation “rule” in question was applied, are perfectly valid. Regarding “proof” 2: ln(2iπ) is of course multi-valued, but then the statement 0=ln(2iπ) is saying that 0 is equal to a whole bunch of different numbers like 2iπ, -4iπ, -8iπ, etc. And the fact that 0 happens to be one among those numbers doesn’t make it any less absurd. Remark: One may also find i^i by Euler’s formula, because it is justified for all complex numbers. Note that complex sine and cosine are defined via Euler’s formula.
I posit there is a gulf between spuriously cancelling digits in a fraction, and unjustifiably applying the third law of exponents, ***in this case***. You state that this law "should [have] some conditions like...", and give the simplest and most restrictive form of those conditions. Wikipedia's "Exponentiation" page gives the "next level up", but once again qualifies it with a disclaimer: "*** In general ***, (b^z)^t b^zt, unless z is real or t is an integer". I suspect the comprehensive expression of those conditions would actually allow it in the case of (e^(i.pi/2))^i
Indeed, I have seen a few videos about complex powers. Every time they start using complex powers without defining what a complex power is, and then applying all the rules of real powers without even checking whether there exists a complex function that has all the same properties as the real power function.
Imaginary/complex numbers in mathematics are like subways, overpasses, bridges, tunnels, etc. They only make sense if they connect something important with something else important. This is the only why they are important in themselves. Otherwise they would be worthless by themselves, they serve to help us move from one "useful" place to another "useful" place.
Root(a) x Root(b) can be written as Root(a×b) if and only if both a,b > 0 (a^b)^c can be written as (a)^bc if and only if a is positive real number. Every Mathematical operation that we know could eventually have a condition that follows
Interesting method the other videos used. At the start of the video I solved for e^-pi/2 by doing i^i = e^ln(i)*i, since e^i*pi/2 = i, e^(i*pi/2)*i = e^-pi/2. Since at the step for substituting the ln has a family of solutions, that is why there are more solutions to i^i
All of the proofs you've shown are fixed by not just including the principal value. For example: 1 = 1^½ = (e^(2πin))^½ ln(1) = ln(e^πin) 0 = πin Which means n=0 Another ln(e^z) z = a+bi ln(e^(a+bi)) a + ln(e^bi) a + ln(e^i(b+2πn)) a + i(b+2πn) 1 = e^2πin ln1 = 2πin 2πin = 0 So n=0 This can be extended into techniques for finding all roots of a number Eg. z³ = 8 => z = ³√8 ³√(8) = ³√(8e^2πin) = 2e^(⅔πin) For n={0,1,2} we get all values for which z³ = 8
5:46 you just simply took the principle value of 1, but in fact for 1=e^(2niπ), n could be ANY integer but not just 1. If you raise 1 to 1/2 power, you should get 1 if n=0 and -1 if n=1. In another way to explain it , 1^1/2=±√ 1. You must figure out all values but not take the principle value and say it is wrong. For (a^b)^c=a^(bc), you may not get all values right so you must take all values
The step doesnt always work, as shown by that particular example, so you cant use it as proof. If you have cases where the step doesnt work and you cant pinpoint the reason, that makes the step unusable if you want to prove something.
@@buycraft911miner2 That's... true. For rings like the real numbers. In complex numbers is usually ok to put equal sign even if something like this happens. This is because most of complex functions are multi-valued.
@@buycraft911miner2 The step doesn’t always work, but at least one of the values is correct. In previous example, 1^1/2=±1, at least one of these is correct. So the step is still useful and intuitive, unlike the 16/64’s example.
That is actually not clear at all, but it can be derived from the Maclaurin series definition of e^z with some sneaky summing. It is probably easier to start with (e^u)*(e^v) and then multiply out their series expansions. From here, collect terms of the same degree and you should see a binomial pattern which can be factored to give something like (u+v)^n in the terms of the series. With a little fudging around in the coefficients, you can get show that you have the exact same series representation as e^(u+v). Since these are unique, they have to be the same function.
Hi, im no expert and didnt understand the video at all, for id like to get an answer to make me understand, why i^i gives such a weird number?, it is just a 1 on a different side of plane, like -1, no matter what it is, is a 1 for example, 1^1, or -1^1 or -1^-1, they will result on a 1 with a symbol that tells on which side of the plane it lays, like, I mean, the number (i) could be rewritten as 1 with another symbol, like $1, so how can it just be that $1^$1 not be either of this four: 1, -1, i, -i?
Both proofs are perfectly valid, both of your demonstrations are wrong. In the first one, 1^½ = ±√1 = ±1 not just 1. In the second, ln(z) refers to the principal log of z, i.e. ln(z) = ln|z| + iArg(z) where -π
Great take-down of people who act as if they know what they're doing in a math video - when they don't - but I'm still a little confused on multiplying i i times.
Presh: e^(iπ/2) = i. My brain: So if I double the exponent on both sides, that gives me e^iπ = i^2, which is -1. Okay, so _that's_ where the Animation vs. Math video got that from. Neat.
6:20 but the square root of 1 (1^0.5) can also be -1 so the answer you get is correct (not the only solution but a correct one). Your example is like saying that you can't use a square root cause you'll get 1 or -1 which means 1=-1. Imo this isn't a satisfying example for proving the problem with using power rules on imaginary and complex numbers.
kind of love on how the answer from the wrong proof's have the same answer as in the real and correct proof, its like: your right for all the wrong reasons
With respect to the faulty wav of doing this, the lesson to be learned is: “It can be easy to get lost when you take a Random Walk on the Complex Plane if you don’t watch where you’re stepping with sufficient care.”
You also forgot to define the argument of a complex number. If you want to define it properly, you can only do so on an open and simply connected subset of the complex plane (we often exclude the negative reel numbers), or else it will not be continuous; I feel like it's a counter intuitive fact people often do not know. This is also required to define properly the log, wheter you want to use the arg function or not.
At 10:45, why does it follow from |z|e^(i arg(z)) = e^u * e^(vi) that |z|=e^u and argz = v? Usually, a*b = c*d doesn't imply that a=c and b=d. It could be that |z|=e^(vi) and argz = -ui, right? Or that |z| and arg(z) are something else completely, like in 3*4 = 2*6.
Note that |e^(ix)| = 1 for real x, and |e^u| = e^u for real u. If you take the absolute value of both sides you get e^u = |z|, implying e^(i arg z) = e^(iv), so that v = arg z + 2πk for integers k.
yeah, considering e^i evaluates to a complex number containing both imaginary and a real number component; his example is like saying, 2(3+i) = 6(1+i/3) therefore 2=6 and 3+i = 1+i/3; which makes no sense. v can equal args z, in which e^u would equal |z|; but, it doesn't really have to, which makes the way he shows this "proof" sketchy. simply put, he omitted the explanation of why and how he was defining: if v = arg z then e^u = |z|.
Tried solving using euler's formula. i^i = exp(ln(i^i))=exp(ilni), then recall exp(ix)=cos x + isinx therefore exp(ilni) = cos(ln(i))+ isin(ln(i)) i= exp(pi/2+2pi*k) but here it breaks and we get result that i^i is complex
Good Video, One Thing that still wasnt explained but was used as I suppose: Ln ( x * y ) = Ln(x) + Ln(y) Where x,y complex Numbers Which is only true for arg(x) + arg(y) < 2 * Pi.
ln(1) as the real logarithm only has one value (it's unfortunate the notation isn't making this distinct, but I think it's a fair criticism that many other videos don't do that)
Saying, "Why every proof you have seen is wrong" in your title is not only arrogant, but you should also know that people living in glass houses shouldn't throw stones.
For proof 2, could it actually be justified? By his explanation, he got 0 = 2iπ, but complex numbers are defined as a + bi, where in this case a = 0 and b = 2π. Wouldn't 0 be defined as a real number in this case, and that "casting" the complex number in the real number system allows for 0 to equal 2iπ? I said 0 is real in this case because 1 is real, and thus ln1 is real.
basically the first two proofs are true for any field of characteristic two provided that we can define the exponential and logarithmic functions (which we can, via the discrete logarithm) in a way that preserves all the properties the exponential and the logarithm have for real numbers
Ok so let's take x a real number, we have 1^x = e^[xln(1)] = e^[x(ln|1| + 2in\pi)] = e^[2inx\pi] which can be egual to any number on the complexe unit circle depending on the value taken for x. Well done for pointing out the issue but I don't get how your way of doing solves it.
Hey Presh. I just came back from a grade 10-12 math contest in McMaster University. There are 12k number of people in a room. They each shake hands with 3k + 6 other people in the room. Any random pair of people you select will have the same number of people who shook hands with both of them that any other pair had people sake hands with both of them. Solve for k.
I feel like if someone is going to go through the trouble of learning the polar form of complex numbers, they are probably doing a full course on complex analysis and would learn about branch cuts. There are certain specific ways that exponent and logarithmic rules for real numbers can apply to complex numbers, but you have to pay attention to the branch cuts. Amusingly enough, I've never encountered these false proofs for i^i because it just doesn't make sense to grab little snippets from complex analysis without talking about the whole picture. That said, nice job exposing your audience to some classic counter-examples on why real number rules don't apply to complex numbers in general as well as deriving the definitions for the power and log functions for complex numbers.
Unfortunately, you missed a nice opportunity to discuss why one needs branch cuts and how one makes functions single valued on a subset of the complex plane. The function you describe for the logarithm is only valid on the Riemann surface of the logarithm, which also yields single valued results, but an infinite number of them given the structure of the Riemann surface. I would say it is incorrect to say we restrict to a principle value unless the branch cut has been discussed. Many branch cuts are possible, which can yield different answers. Understanding how you restrict the complex plane to make the results into continuous and even analytic functions is the important result when raising a complex number to a complex power.
There is nothing wrong about -1 being the square root of 1. We did not say that 1 is the only answer to being the square root of 1. Therefore, it is wrong to say 1 = -1, since 1 could also be exp(4πi), in which case it's 1/2 power is indeed 1.
For the second ”proof“, since you have taken logs of both sides, we are essentially looking at the arguments, in which case, 0 is indeed the same as 2π. Cos you can just keep going and say 2iπ = 4iπ = 6iπ etc etc.
8:04 "0 = 2*i*pi" - is there any significance that in terms of angles, 0 rad and 2pi rad are equivalent, and does 2*i*pi make it any different in terms of angles?
The correct way is thankfully the only way I ever learned to do complex exponentiation. What I really want to know is how you came up with those faulty fraction cancellation examples!!!
The proofs I've seen have been more of a "checking your work." They aren't true proofs as far as I'm concerned. i intuitively shows rotation about an original but you need additional details about the i to determine if it itself is even rotating. So some of the equations with i look like they'd make more sense if the output i was upside down or to the side or inside out but people don't take it that far. The proofs I've seen is like saying 1+1=2 because 2-1=1. That's more of a proof of correlation.
What you said in the introduction is wrong by the way. Complex numbers were a byproduct of solving for (real) zeroes of polynomial equations of degree 3 and higher
I really like the canceling digits method example 😂
Truly😂😂
It’s called anomalous cancellation for those who are curious
It killed me
I like using them in my math class to illustrate how the wrong method can still get the right answer on occasion, and show the common error of improper "cancellation" to simplify fractions.
Just came by to drop x^2 = 25
See ya
At 6:12, the reason you get a nonsense answer here is independent of the (a^b)^c = a^(bc) step. You've stated that 1^(1/2) = 1, which is correct, used (a^b)^c = a^(bc) to find that 1^(1/2) = -1, which is also correct. The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another.
If you are careful to always note that e^f(x) = e^(f(x) + 2 i n pi), where n is an integer, then that avoids the problems associated with (a^b)^c = a^(bc). Of course, using this method isn't helpful when the exponent has a base of 1, because using a base of 1 is generally problematic and breaks a lot of general rules. For example, 1^(infinity) is undefined rather than infinite.
There are other examples of (a^b)^c = a^(bc) not holding for complex numbers. The point in the video is still true.
1^(1/2) = 1 It doesn't have 2 solutions, you're mistaking it with x^2=1 which would be |x|=1^(1/2) the x is inside a module so there you have 2 solutions.
@@Max_Power_ That only holds for real numbers. For complex numbers, functions can be multivalued.
@@Max_Power_ No, you're thinking of sqrt(1), which is positive by definition
"The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another." - ya, because the sqrt function has 2 branches, one of which is the principle one
(a^b)^c is indeed not always applicable to complex numbers, however, your example is *wrong*.
1 = e^2PI*i
then you raise both sides to the power of 1/2, effectively taking the square root. The square root function is not injective, you should have added +- to one of the sides, which would have made the equation correct.
Basically you've shown that both 1 and -1 are square roots of 1 (which is not a mistake)
Agree, but in my opinion, (a^b)^c is applicable in case of purely 1 dimensional calculations. For example, here, We are purely using y-axis, so using this formula gives correct answer.
When it comes to 2D calculations, like vectors (in complex number when both y axis and x axis are involved), this formula should become invalid, or should be generalised to get proper answer.
@@raffaeleraia2223 The thing is, square root of 1 truly have two values 1 and -1. But, we defined it as only positive 1.
In the equation it will always give two values.
And, a mathematical solution doesn't necessarily follow what we defined, unless we change the solution to make it follow what we defined.
I am not good at words here it seems.
did presh get back to you on this point of adding +- on both sides since we are dealing with the square root?
@@imfuture7543the reason square root usually gives a positive and not a negative is due to the fact that having just a single output is useful. so to do that the principal value is given
Why does being injective matter ? The square function is also not injective and 1 = x => 1² = x² is still true. Adding +- would be extremely wrong too since x -> x^½ is a single value function
it depends on how you define complex logarithms. if you want to have a function, you need to allow a principal value of the logarithm. if you want to find all solutions, it isn't a normal function anymore because it has multiple outputs. method 2 isn't a "wrong" proof, it just glosses over some small detail with +2*k*pi*i in the logarithm function which does cancel out when you raise it to e. It's obviously different from the fraction coincidence at the beginning because this is a good method that always works for any similar problem. you can proof euler's theorem with taylor series or you could just assume the result. doesn't make it wrong.
I'm a wrong believer in two wrongs make a right. Let me repeat that, I'm a wrong believer in two wrongs make a right.
Actually the 2kπi won’t cancel out because you have an extra i, so raising it to e would give you a factor of e^(-2kπ), which contributes to the multi-valued nature of i^i itself.
@hapedisedivide1980 What are you sayin' ?
@@samueljehanno Please simply refer to the video and compare.
There is nothing wrong with the first two proofs.
Those are functions that give multiple branches, which is why you can produce an apparent contradiction, but in each case the calculation does still give _A_ branch, and that branch is a valid evaluation.
However, choosing a specific branch without giving a reason why, is still wrong. The first proof in fact technically falls into the fallacy mentioned with canceling digits - for example, canceling digits works as long as the number of trailing zeros in the numerator is greater than or equal to the number of trailing zeros in the denominator, and the denominator is a power of 10, and the only digit being canceled is 0. But still it's an unjustified way to prove for the others.
The second one is more like you're saying, but still, choosing a branch requires justification, otherwise you can never be sure that you chose the correct branch. So it is technically wrong (but incomplete is a better description)
@@hbrg9173That's basically the principal branch, and taking the principal branch by default is reasonable to me.
I would say these proofs are to introduce the concept of index form rather than slam dunking the audience terms of complex analysis which is complex as u see by its name.
If there are multiple branches the function isn't bijective though.
@@d3ds1r There are two topics. For solving equation, it's totally fine to have multiple answers.
But if you want to define a function, You usually apply a polar form and construct a Riemann Sphere to try to make an only output.
This works more easily if you use the unwinding number.
Your counterexample to proof 1 is finding out that 1 has two square roots, one of which is 1, and the other of which is -1. This is not wrong. That's always something you need to consider when undoing a root with a square! You're focusing on an extraneous root.
In general though, the core of this is that when you're using multivalued complex functions, yes, you do in fact need to be careful because the principal value isn't always the one you're looking for! This is why the proof 2 got the nonsensical result of 0 = 2pi*i
So yes, the careful analysis is accurate.
@@sylverfyreThe simplest solution is to note that polar representations are not unique. The issue is not the algebraic operation in the exponent as he makes it seem. That is perfectly fine. And roots are taken care of by applying De Moivre’s formula. (Actually even complex roots are no trouble since ℂ is a field and we can always rewrite a rational function in i as a linear polynomial in i.)
with complex numbers, the log function and fraction exponents are multivalued. in the steps where you "disproved" rules of math for complex numbers you conveniently decided to ignore the multivalued nature of these operations, which you can not do.
The video is a nice explanation of branch cuts and all, but I think a lot of it can be avoided by simply noticing that a complex number has more than one polar representation. Then the first “proof” works trivially.
I mean, ℂ is a characteristic 0 field for crying out loud. That’s a pretty nice algebraic structure as far as I’m concerned.
Thank you, I was getting so mad watching those "disproofs." Conveniently ignoring the multivalued nature of roots and logarithms to say things are wrong.
Literally every video (and in english like he mentions) I've seen showing i^i is real either explains that they're taking principle values of everything meaning that ln(e^2ipi) would _indeed_ be 0, or they specifically mention that it's multivalued and get the final result with n somewhere indicating its multivalued nature.
@@just_a_dustpan Principal values, not principle values.
I love complex analysis! In the case of the natural logarithmic function, we can make a branch cut and define the principal branch as log z = ln r + iθ, where r > 0 and θ is in (-π, π]..
How would you define log(-1) then? It has no argument in the range you stated.
@@MichaelRothwell1 I erroneously excluded π because I couldn’t find a
Cutting a locally holomorphic function kinda defeats the pupose of it being locally holomorphic in the first place. Path dependence is just the way to go in complex analysis. If cutting is required, you would want to choose the cut freely anyway.
My favourite aspect about mathematics is that any two people independently discovering the same subject should derive equivalent definitions.
@@caspermadlener4191 In this case, the reason why we define a branch cut is because the logarithmic function is multiple-valued.
@@JaybeePenaflor Thanks for the clarification! (here is the ≤ symbol, in case you need it in the future).
I mean, the second proof was basicaly showing that e^(-pi/2) is an answer. The only thing it did differently from the final proof was considering just 1 answer to ln(x), but this answer is still valid so it should also get only one answer to i^i (and it did)
doesnt the 1=-1 proof fail because you decided that
squareroot 1 = 1
instead of
squareroot 1 = +-1 ?
For complex numbers, if you account for all branches the identity (a^b)^c=a^(bc) should hold.
likewise in proof 2 counterexample, the natural log needs to be defined respective to a branch of the natural log which effectively reduces it mod 2πi. if you want to avoid that you need to choose a branch which includes 0 properly, so you can let πi =-πi and then you wouldnt get 0=2πi rather 0=0
(a^b)^c=a^(bc) does hold if you treat it as a multivalued function. But I think the point in the video is that typical presentations do not do that.
@@TheEternalVortex42 That's actually not true, although Presh's example doesn't make that clear. There is an example due to Clausen where the formula fails, even when considering the multi-valued case. (See the Wikipedia page "Exponentiation" and search for "Clausen").
@@TheEternalVortex42I agree that that is the point of the video. But I disagree that that is an actual typical problem. I feel like he was assuming presentstions have the problems he's saying, but they just don't.
Yep. Squareroot is a multi-valued function, which results in exactly the same problem that taking the log has.
When, in "proof" 1, you substitute 1^(1/2) for 1, you must also say that 1^(1/2) has two results, namely 1 and - 1. So the power law holds for one of the values.
Exactly. It's a multivalued complex root/power.
I've seen "log" used for the complex log function along with "ln" for the real natural logarithm.
No, "log" is used for log-base-10 (though some weirdos at ISO have decided that "lg" means base-10 when that is more commonly used for base-2...the idea is that "log" means the generic logarithm where you need to provide the base, making the binary logarithm "lb.")
Are you thinking of "Log", with a capital L?
@@dhwyll In the Wikipedia entry for complex log, they use "log" as a generality, and "Log" for the principle value.
@@dhwyllit depends on the field of study. In some texts that focus on graph theory and computer algorithms, they write log and it is assumed to be based 2. In some complex analysis texts, log is base e. You have to adopt the convention of the author and the field of math you are working in.
@@dpgoulette Exactly 😊
At high school, we used the names ln, log and log_a, for e-based, 10-based and a-based logarithms.
At the university we used (almost) only the natural logarithm and named it log.
That is: Check the definition of log in the text you are studying, and if the definition is not provided you have a natural choice 😂.
@@dhwyllAnd as we all know, the generic logarithm is the logarithm with base e, aka. natural log.
Dam this definitely blew my mind🤯
I have been so used to doing these problems normally that I almost forgot the limits of the basic mathematics rules like this indices one!
The multi valued result is still ridiculous. It means that all these values are equal to i^i, but not equal to each other. This violates the postulate that if a=b and b=c the a=c.
8:00 The natural log of any number except zero has infinite solutions. When you take the ln of 1, it would give the following solutions: ln(1) = 0 + n*2*pi*i, where n is an integer. Similar to ln(e^(2*pi*i)), it would result in: ln(e^2*pi*i) = 2*pi*i + n*2*pi*i = (n+1)*2*pi*i = m*2*pi*i, where m is an integer. This means: ln(1) = ln(e^(2*pi*i)).
*Yeah so that's basically the same as saying "2 = -2 because they both square to 4, but that's impossible, so we can't define square roots."*
@@mr.d8747the nth root FUNCTION is defined using the principal branch. Therefore, only 1 value comes from it.
The 1/n exponent takes all branches. It's a multivalued function and you have to specify the branch
presh sounds so done with everything
Than explains so much! Thank you! No matter if it's i^i or whatever else, the fact you made me pay more attention while carrying out supposedly obvious operations is priceless.
you have taken 2nd principal value for 1 which will give you -1 its no surprise there as 1^1/2 may equal to -1. but for 1=e^0 1st principal value it will remain 1. Well new method is also correct but the previous method also works we just need to be careful. as we have applied the the method to find roots of unity x^5=1.
Presh, you're usually right, but this time... All the values of the "counter-proofs" are a bit of fraudulent.
In the first, you took not the principal value of one in the complex plane but another one (1=e^2πi) and you stated that is equal to one (true). But then you took the square root. 1 is the principal complex square root of the complex number 1, but you took the complex square root (which is multi-valued) of e^2πi. In other terms, you took the real square root on the LHS (which matches with the principal complex square root on complex numbers with imaginary part = 0) and the complex square root (multivalued) in RHS. If you treated 1 with the general argument (e^i(0+2πk), k is an integer) you wouldn't get to this result because it has another value which is 1.
In the other, happens the same thing. You took the real natural logarithm on the LHS (which matches the principal branch of the natural logarithm on complex numbers with imaginary part = 0) and the complex natural logarithm on the RHS (the multi-valued natural logarithm). That property has to be true, not also because you use it when solving the equation z = e^w, but because is the definition of the natural logarithm (and it has to be multi-valued because so it is the complex exponential)
In both cases, you took two different functions on the two sides, making the answer blatantly wrong. In complex numbers, is better to work with the general argument and then reduce it to the principal form if you're not sure what you're doing.
Technical misunderstandings: Ln(z) is the notation of principal branch of natural logarithm, usually not the notation for real natural logarithm. It's subtle because for real numbers is equal, but not for complex numbers. Use Log(z) and log(z) notation for complex natural logarithm (since defining based-logarithms is a bit useless in the complex world) and ln(x) for natural real logarithm.
Thank you for reading to anyone who kept reading so far and I love you Presh, this is just constructive criticism
Slight correction on proof 2:
Since e^{2i\pi} = e^{0}, and in fact any angle shifted by 2\pi is the same angle on the unit circle, it can be said that 0 is in fact equal to 2i\pi. Or more precisely, this is a number system in which all numbers in the exponent are modulo 2\pi, and mod(2\pi, 2\pi) = 0.
It is still interesting to see where these steps can go wrong, so thank you for the informative video.
the step at 13:25 states that the modulus of i is equal to 1 but when we will open the modulus we get :
i = ± 1 that means √-1 = ± 1 which is INCORRECT!
Isn't this the method that Blackpenredpen uses? So you wouldn't be the only one that uses a justified method.
Putting 1 as √1 is slightly deceptive (Though theoretically correct) since there are two numbers whose square is 1 I e 1 and -1
he didn't put 1 as root(1), he put root(1) as 1, not the same thing
The whole cancelling digits thing is really interesting. I want to know what process can lead you to find a general solution for what fractions that happens to
Brute force of Chuck Norris 😅
This has to be one of your best ever videos, if not the best - bravo!
Or, you could just do it directly. The formula for complex exponentiation is:
(a + bi)^(c + di) = (a^2 + b^2)^[(c + di)/2] * e^[i(c + di) * arg(a + bi)]
Where arg(a + bi) is the angle with the positive real axis for the number.
Thus, for i^i, that is (0 + i)^(0 + i). Therefore, a = 0, b = 1, c = 0, d = 1. Thus:
(0 + 1i)^(0 + 1i)
(0^2 + 1^2)^[(0 + 1i)/2] * e^[i(0 + 1i) * arg(0 + 1i)]
1^(i/2) * e^[i(i) * (π/2 + 2πn)]
1 * e^-(π/2 + 2πn)
e^-(π/2 + 2πn)
Thanks! Where did you get this formula?
@@nitey123 My Complex Analysis class. It's a consequence of Euler's formula and the polar coordinate nature of the complex plane.
Sorry, got lost in the middle of the whole thing. IMHO you also fell in the pitfall you wanted to avoid, and this started the moment you presented i in polar form. The real problem, before you get to i^i is to explain, what it means to elevate anything to a power of a complex number.
What I mean is that sometimes the meaning of some mathematical operators become a little esoteric. For instance a+b is a + 1 + 1 + 1 ...+1 (b times) a * b is a + a + a ... (b times) +a. But when it get to exponentiation, and we start having a^1/2, it starts being something else than a * a (half a time?). The notation starts taking a special meaning. And this is a simple point that could be better explored in the video.
It's amusing that all the comments point out that the counterexamples are only "wrong" because they ignore that complex log is multivalued. Yet the whole point of that part of the video is that other videos are ignoring that fact!
This would be great if Presh included the part about e^z also being multivalued and different from exp(z)
A standard example of the “Cancel the Digits” technique:
Given y = sin x / n
Cancel the n in the numerator and the denominator
Resulting y = 6
At 6:00 it was you that messed up. 1 = e^(2*pi*i) you were fine, but when you took the square root of both sides you messed up. -1 is also equal to 1^(0.5). You played a trick with the multi-valued nature of the square root function. The issue has nothing to do with the "multiply the exponents" rule - it was another mistake altogether.
To put it another way, when you introduced the 1/2 exponent, you took the principle value on the left but the other value on the right.
presh still amazes his audience with his mastery or mathematics both on the elementary and advanced levels: presh, once again, challenges me to catch-up and review mathematic principles: touché presh
I see a lot of people saying that he ignored +/-. He did not. He is using the function definition of sqrt, which is what you normally use for proofs. Using the operator definition is only fine for solving an equation (in which case you have to test for pseudo-answers). The reason the operator definition can't be used in proofs is because sqrt(1)=+1 and sqrt(1)=-1 is true in an equation, but in a proof, this would mean that per the transitive property of equality 1= -1, which is not true.
Technically speaking if we look at negative 1 and positive 1 as being unit vectors, their magnitudes are absolutely equal, it's just that their directions are point away from each other along the same line. The radian measure from 1 to 1 is + / - PI radians or + / - 180 degrees. The arccos of the dot product between the points (1,0) and (- 1,0) through the definition of the cosine function is PI radians or 180 degrees. So in some ways 1 and - 1 are kind of equal. This is why we can write them as + / - 1 which is referred to plus or minus one. The operative logical word here is OR. This is why the absolute value function always returns a positive value for all non zero numbers. If we are concerned with the actual forward direction of the vector then the above statement wouldn't be + / - 1, plus or minus one, instead it would be plus AND minus one where AND would return false. Within the first context of plus OR minus, if either is true then the entire statement is true.
How is this possible? How can - 1 and + 1 simultaneously be equal to while not being equal to each other? Well think of | + / - 1| as being an arbitrary unit vector. Stand in place facing north and reach your arm out in front of you. Your looking direction is forward. The length of your arm being an arbitrary unit of measure is worth 1 unit of your arms length. Now while keeping your are stretched out in front of you turn 180 degrees or PI radians to face south. Your arm being a unit length away from your body being the center or focal point is the radius of a half circle. Your arm created an arc, a curved line. The radius of this arm, the length of your arm did not change. It is equivalent the entire time. What has changed was your facing direction. You were pointing north, and now your are point south. The points at due north and due south make either a straight line, or a curved geodesic if on a multi dimensional curved surface such as a sphere or a cylinder. This is why polar coordinates work in the way they do, this is why the complex numbers exhibits the properties of rotations and they are easily convertible. The value doesn't change, the direction does. Even within every day use such as with currency with credit and debt... If someone gives you a dollar you gained a dollar and you are +1 in the books. If you owe a dollar to someone else to where you have to give it away, you are at a loss and are - 1 in the books. The value of the dollar is still 1. The direction it is going is what changed.
These directional differences between 1 and - 1 are parallel directions. They're not orthogonal. Orthogonal would be 1/2 of that rotation or 1/2 of that direction. Orthogonal directions would be East and West or Up and Down from North and South. This is why the multiplication of i is equivalent to a 90 degree or PI/2 radian rotation. Yes, we've been calling them imaginary and complex numbers for years... In truth I think this nomenclature is truly an inaccurate description. The complex terminology is just fine the way it is because the complex numbers are just that, they are complex because there is a real part and there is a secondary part that we have been calling imaginary. I think what ought to be changed, but probably won't happen because of the billions invested in writing all of the text books... would be to accurately call the imaginary numbers the orthogonal or perpendicular numbers. That's what they are. They are orthogonal - perpendicular to the real numbers. They are rotated 90 degrees or PI/2 radians from the real number plane. The square root, well in fact, the even roots of all negative numbers are orthogonal to the reals. This is why there are even and odd functions both within the polynomials as well as the trigonometric functions. x^2, x^4, x^6, are even and x, x^3, x^5, etc... are odd. The cosine and secant are even functions, and the sine, cosecant, tangent and cotangent are odd functions. Yet the sine and cosine functions have the same exact wave form, same period, same range and domain, the same properties of their limits, the only difference is their initial starting positions, The sine starts at (0,0), and the cosine starts at(0,1). The points (0,0) and (1,0) are 90 degrees from each other. This makes a vertical line which is perpendicular to the x-axis or the domain of the given function. Also their graphs are 90 degree or PI/2 horizontal translations of each other, they are 90 degrees out of phase.
If we dive a bit deeper we can see that both the Pythagorean Theorem and the equation of the circle for most tense and purposes are the same thing, it's just that the Pythagorean Theorem is within the context of Right Triangles and the Equation of the Circle relates the center of it's circle, it's radius, and the arc that it generates. This is why the trigonometric functions have Pythagorean Identities. When we dive in to Physics especially within the realm of electricity or the electromagnetic spectrum, the real and "imaginary" parts or what I like to call the orthogonal parts of the waves functions is quite intriguing. This can also be seen within optics. It pertains to wave propagation. So if we now facing South and rotating in the same turning direction as before and rotate another 180 degrees or PI radians... We are now facing north again. You are pointing in the original direction, you haven't displaced yourself (horizontal linear transformation of translation), you only rotated in place, and the length of your arm hasn't changed. We have just went full circle, and this is why things such as reflections and symmetry exist. Think of + 1 and - 1 as being orthonormal reflections of each other meaning that they are pointing away from each other along the same line of sight from a common center point. If they were facing the same directions, the arccos would 0 as in 0 radians or 0 degrees (here we don't included 2PI or 360 because of the range and domain of the arccos function, and if they were pointing towards each other then the arccos would then produce? I think this would be another complex number...
So that's how two values can simultaneously be equivalent yet not equivalent to each other. They have the same magnitude or absolute value but their sign, direction, or heading is different. And what's quite interesting is if you graph both the arccos and arcsine functions together it draws the outline of an hour glass, and it also takes on the representative shape of the double helix or the structure of DNA...
Saw some people disagreeing with the argument against the “proofs”, so here’s a bit of a take of mine.
Regarding counterexample in “proof” 1: The exponential function is single-valued, in fact holomorphic, on all of C. Therefore all the steps in the counterexample, except where the erroneous Exponentiation “rule” in question was applied, are perfectly valid.
Regarding “proof” 2: ln(2iπ) is of course multi-valued, but then the statement 0=ln(2iπ) is saying that 0 is equal to a whole bunch of different numbers like 2iπ, -4iπ, -8iπ, etc. And the fact that 0 happens to be one among those numbers doesn’t make it any less absurd.
Remark: One may also find i^i by Euler’s formula, because it is justified for all complex numbers. Note that complex sine and cosine are defined via Euler’s formula.
I posit there is a gulf between spuriously cancelling digits in a fraction, and unjustifiably applying the third law of exponents, ***in this case***.
You state that this law "should [have] some conditions like...", and give the simplest and most restrictive form of those conditions.
Wikipedia's "Exponentiation" page gives the "next level up", but once again qualifies it with a disclaimer:
"*** In general ***, (b^z)^t b^zt, unless z is real or t is an integer".
I suspect the comprehensive expression of those conditions would actually allow it in the case of (e^(i.pi/2))^i
blackpenredpen gave the correct answer using a "right" method 6 years ago
Indeed, I have seen a few videos about complex powers. Every time they start using complex powers without defining what a complex power is, and then applying all the rules of real powers without even checking whether there exists a complex function that has all the same properties as the real power function.
Imaginary/complex numbers in mathematics are like subways, overpasses, bridges, tunnels, etc. They only make sense if they connect something important with something else important. This is the only why they are important in themselves. Otherwise they would be worthless by themselves, they serve to help us move from one "useful" place to another "useful" place.
But the supposed error at 7:56 can easily be remedied by adding 2ipi Plus 2npi Presh so technically it's not all wrong..wouldn't you agree
The lesson here is: when in doubt trust Wolfram not RUclips.
finally someone does it in right way
I think the thumbnail is wrong. You said i^i = e^pi/2, not i^i = e^-pi/2.
For me thumbnail is showing "?" 😂
Root(a) x Root(b) can be written as Root(a×b) if and only if both a,b > 0
(a^b)^c can be written as (a)^bc if and only if a is positive real number.
Every Mathematical operation that we know could eventually have a condition that follows
for proof 2, it’s also correct. The key point is that i=e^i(π/2+2kπ), then everything is correct. 6:43 is wrong on purpose i think.
Interesting method the other videos used. At the start of the video I solved for e^-pi/2 by doing i^i = e^ln(i)*i, since e^i*pi/2 = i, e^(i*pi/2)*i = e^-pi/2. Since at the step for substituting the ln has a family of solutions, that is why there are more solutions to i^i
As always I can relax watching your videos, because you explain strict mathematics in a very pedagogical way 😊.
All of the proofs you've shown are fixed by not just including the principal value.
For example:
1 = 1^½ = (e^(2πin))^½
ln(1) = ln(e^πin)
0 = πin
Which means n=0
Another
ln(e^z)
z = a+bi
ln(e^(a+bi))
a + ln(e^bi)
a + ln(e^i(b+2πn))
a + i(b+2πn)
1 = e^2πin
ln1 = 2πin
2πin = 0
So n=0
This can be extended into techniques for finding all roots of a number
Eg.
z³ = 8 => z = ³√8
³√(8) = ³√(8e^2πin) = 2e^(⅔πin)
For n={0,1,2} we get all values for which z³ = 8
Love that you would make a 15 minute video on this, and I would be absorbed watching it all. Nice job.
5:46 you just simply took the principle value of 1, but in fact for 1=e^(2niπ), n could be ANY integer but not just 1. If you raise 1 to 1/2 power, you should get 1 if n=0 and -1 if n=1. In another way to explain it , 1^1/2=±√ 1. You must figure out all values but not take the principle value and say it is wrong. For (a^b)^c=a^(bc), you may not get all values right so you must take all values
The step doesnt always work, as shown by that particular example, so you cant use it as proof. If you have cases where the step doesnt work and you cant pinpoint the reason, that makes the step unusable if you want to prove something.
@@buycraft911miner2 That's... true. For rings like the real numbers. In complex numbers is usually ok to put equal sign even if something like this happens. This is because most of complex functions are multi-valued.
@@buycraft911miner2 The step doesn’t always work, but at least one of the values is correct. In previous example, 1^1/2=±1, at least one of these is correct. So the step is still useful and intuitive, unlike the 16/64’s example.
@@KewoNg-to6zj yeah, I do agree that, although funny, the example isnt even close to the same case
At 8:50, how did you come to that conclusion/derivation for the complex exponent? Seems like that was not well justified?
That is actually not clear at all, but it can be derived from the Maclaurin series definition of e^z with some sneaky summing. It is probably easier to start with (e^u)*(e^v) and then multiply out their series expansions. From here, collect terms of the same degree and you should see a binomial pattern which can be factored to give something like (u+v)^n in the terms of the series. With a little fudging around in the coefficients, you can get show that you have the exact same series representation as e^(u+v). Since these are unique, they have to be the same function.
There are always N Nth roots. So this does not work, (-2)^2 = 4 = 2^2. Therefore -2 = 2.
Hi, im no expert and didnt understand the video at all, for id like to get an answer to make me understand, why i^i gives such a weird number?, it is just a 1 on a different side of plane, like -1, no matter what it is, is a 1 for example, 1^1, or -1^1 or -1^-1, they will result on a 1 with a symbol that tells on which side of the plane it lays, like, I mean, the number (i) could be rewritten as 1 with another symbol, like $1, so how can it just be that $1^$1 not be either of this four: 1, -1, i, -i?
Both proofs are perfectly valid, both of your demonstrations are wrong. In the first one, 1^½ = ±√1 = ±1 not just 1. In the second, ln(z) refers to the principal log of z, i.e. ln(z) = ln|z| + iArg(z) where -π
10:19 Here you said "since u and v are real, you can say e^(u+vi) = e^(u)e^(vi)"
Why does the i not make a difference?
Great take-down of people who act as if they know what they're doing in a math video - when they don't - but I'm still a little confused on multiplying i i times.
Presh: e^(iπ/2) = i.
My brain: So if I double the exponent on both sides, that gives me e^iπ = i^2, which is -1. Okay, so _that's_ where the Animation vs. Math video got that from. Neat.
6:20 but the square root of 1 (1^0.5) can also be -1 so the answer you get is correct (not the only solution but a correct one). Your example is like saying that you can't use a square root cause you'll get 1 or -1 which means 1=-1. Imo this isn't a satisfying example for proving the problem with using power rules on imaginary and complex numbers.
I am encouraged that there are still people in this world that consider these discussions entertaining.
At age 80 I find these discussions entertaining as a way of staving off dementia.
kind of love on how the answer from the wrong proof's have the same answer as in the real and correct proof, its like: your right for all the wrong reasons
1=X^2 simplfies to X=1^(1/2). Which then simplifies to X can be either 1 or -1.
With respect to the faulty wav of doing this, the lesson to be learned is:
“It can be easy to get lost when you take a Random Walk on the Complex Plane if you don’t watch where you’re stepping with sufficient care.”
Same answer but that is the best proof i've seen so far. Keep up the good work.💯💯
i^i= e^(-pi/2) for all values for ln(z), not just the principal.
You also forgot to define the argument of a complex number. If you want to define it properly, you can only do so on an open and simply connected subset of the complex plane (we often exclude the negative reel numbers), or else it will not be continuous; I feel like it's a counter intuitive fact people often do not know. This is also required to define properly the log, wheter you want to use the arg function or not.
At 10:45, why does it follow from |z|e^(i arg(z)) = e^u * e^(vi) that |z|=e^u and argz = v?
Usually, a*b = c*d doesn't imply that a=c and b=d.
It could be that |z|=e^(vi) and argz = -ui, right? Or that |z| and arg(z) are something else completely, like in 3*4 = 2*6.
|z| is always a real number.
Note that |e^(ix)| = 1 for real x, and |e^u| = e^u for real u. If you take the absolute value of both sides you get e^u = |z|, implying e^(i arg z) = e^(iv), so that v = arg z + 2πk for integers k.
yeah, considering e^i evaluates to a complex number containing both imaginary and a real number component; his example is like saying, 2(3+i) = 6(1+i/3) therefore 2=6 and 3+i = 1+i/3; which makes no sense. v can equal args z, in which e^u would equal |z|; but, it doesn't really have to, which makes the way he shows this "proof" sketchy. simply put, he omitted the explanation of why and how he was defining: if v = arg z then e^u = |z|.
Great job with the details, and why the other proofs were incorrect.
Tried solving using euler's formula.
i^i = exp(ln(i^i))=exp(ilni), then recall exp(ix)=cos x + isinx therefore exp(ilni) = cos(ln(i))+ isin(ln(i))
i= exp(pi/2+2pi*k) but here it breaks and we get result that i^i is complex
I wonder because when we write 1 as e^0i then proof 1 works but when we write 2π it doesn't, and 2π is actually 0 only?
Just take log
Good Video,
One Thing that still wasnt explained but was used as I suppose:
Ln ( x * y ) = Ln(x) + Ln(y)
Where x,y complex Numbers
Which is only true for
arg(x) + arg(y) < 2 * Pi.
The first example should be reworked because the square root of one can also be negative one, so technically it is correct
7:55 ln(1) can also be 2πi, 4πi, kπi. I don't see the problem here.
ln(1) as the real logarithm only has one value
(it's unfortunate the notation isn't making this distinct, but I think it's a fair criticism that many other videos don't do that)
If you accept the Ramanujan Summation, you HAVE TO accept i=e^(i•pi/2)
Fantastic! Could you please do this with Ramanujan's sum -1/12? Please!!
most videos showing that 1+2+3+... = -1/12j ust assume a series to converge when it isn't. That's it.
Actually sum of all numbers is not truly -1/12 imo. But, using it often solves the problem, since you can't always deal with infinity.
The thumbnail is wrong. It is written π/2 rather than -π/2 😅.
Yep, I could not believe he actually got the thumbnail wrong. I looked at it twice and then a third time, figuring, how could he get that wrong?
Saying, "Why every proof you have seen is wrong" in your title is not only arrogant, but you should also know that people living in glass houses shouldn't throw stones.
For proof 2, could it actually be justified?
By his explanation, he got 0 = 2iπ, but complex numbers are defined as a + bi, where in this case a = 0 and b = 2π.
Wouldn't 0 be defined as a real number in this case, and that "casting" the complex number in the real number system allows for 0 to equal 2iπ?
I said 0 is real in this case because 1 is real, and thus ln1 is real.
basically the first two proofs are true for any field of characteristic two provided that we can define the exponential and logarithmic functions (which we can, via the discrete logarithm) in a way that preserves all the properties the exponential and the logarithm have for real numbers
Ok so let's take x a real number, we have 1^x = e^[xln(1)] = e^[x(ln|1| + 2in\pi)] = e^[2inx\pi] which can be egual to any number on the complexe unit circle depending on the value taken for x. Well done for pointing out the issue but I don't get how your way of doing solves it.
What an elegant video!
"Everybody would object using this method".
Sir, you are underestimating me xD.
Hey Presh. I just came back from a grade 10-12 math contest in McMaster University. There are 12k number of people in a room. They each shake hands with 3k + 6 other people in the room. Any random pair of people you select will have the same number of people who shook hands with both of them that any other pair had people sake hands with both of them. Solve for k.
why is the absolute value of i equal to 1 ??
Is this a default identity, the same way that x^0 is ALWAYS 1 (ONE) ?? (assuming x is a natural number)
.
Neither abs(i) =1, nor x^0=1.
I feel like if someone is going to go through the trouble of learning the polar form of complex numbers, they are probably doing a full course on complex analysis and would learn about branch cuts. There are certain specific ways that exponent and logarithmic rules for real numbers can apply to complex numbers, but you have to pay attention to the branch cuts. Amusingly enough, I've never encountered these false proofs for i^i because it just doesn't make sense to grab little snippets from complex analysis without talking about the whole picture. That said, nice job exposing your audience to some classic counter-examples on why real number rules don't apply to complex numbers in general as well as deriving the definitions for the power and log functions for complex numbers.
False, polar form is used often in circuit analysis - the classes at least. Very few of those engineers are going to take any kind of analysis course.
@@cokesucker9520 I pity your EEE students, ours learn complex analysis at 4th semester
"i" like your videos.
Unfortunately, you missed a nice opportunity to discuss why one needs branch cuts and how one makes functions single valued on a subset of the complex plane. The function you describe for the logarithm is only valid on the Riemann surface of the logarithm, which also yields single valued results, but an infinite number of them given the structure of the Riemann surface. I would say it is incorrect to say we restrict to a principle value unless the branch cut has been discussed. Many branch cuts are possible, which can yield different answers. Understanding how you restrict the complex plane to make the results into continuous and even analytic functions is the important result when raising a complex number to a complex power.
This video answered my so many questions
There is nothing wrong about -1 being the square root of 1. We did not say that 1 is the only answer to being the square root of 1. Therefore, it is wrong to say 1 = -1, since 1 could also be exp(4πi), in which case it's 1/2 power is indeed 1.
For the second ”proof“, since you have taken logs of both sides, we are essentially looking at the arguments, in which case, 0 is indeed the same as 2π. Cos you can just keep going and say 2iπ = 4iπ = 6iπ etc etc.
"There is nothing wrong about -1 being the square root of 1" yes there is. Because it isn't.
@@salerio61it is the square root but not the principal root
I have learnt complex number recently so I now I can understand these proof
Squaring usually introduces extraneous roots.
for proof n2
didn't we say e^0i=e^2πi? then obviously it will come out that 0i=2πi because θ is cyclical
8:04 "0 = 2*i*pi" - is there any significance that in terms of angles, 0 rad and 2pi rad are equivalent, and does 2*i*pi make it any different in terms of angles?
The first counter example is wrong.
You basically said: 1=1^1/2
-1=1^1/2 (which is true)
Which means 1=-1
Thats like saying if a²=b² then a=b
The existence of e^ipi = -1 cancels the entire rant about logarithms lol
12:25
0
The correct way is thankfully the only way I ever learned to do complex exponentiation.
What I really want to know is how you came up with those faulty fraction cancellation examples!!!
The proofs I've seen have been more of a "checking your work." They aren't true proofs as far as I'm concerned. i intuitively shows rotation about an original but you need additional details about the i to determine if it itself is even rotating. So some of the equations with i look like they'd make more sense if the output i was upside down or to the side or inside out but people don't take it that far. The proofs I've seen is like saying 1+1=2 because 2-1=1. That's more of a proof of correlation.
What you said in the introduction is wrong by the way. Complex numbers were a byproduct of solving for (real) zeroes of polynomial equations of degree 3 and higher
But why do the wrong methods work?