@don'tfeel I’m curious, and as this question shows I am not very knowledgeable: how is the fact that i^i is equal to a real number related to hyperbolic geometry?
@@PunmasterSTP Joke aside, its actually not hard to imagine at all. 4^(1/3) both of these numbers are rational the output is irrational but it's fine because both rational and irrational numbers are Real similarly, both imaginary and real numbers are Complex!
@@wraithlordkoto Oh I quite agree; I just try to sneak in puns where I can 😎 I also think it’s interesting how one irrational number can be raised to the power of another irrational number, and end up being an integer!
@@shaunthedcoaddict1656 look, every complex number z = x + iy can be represented as it's magnitude |z|(which is a real value), times it's "rotation" which is expressed as e^{i*theta}, where theta is the angle the line from 0,0 to the point z in the argand plane, or tan theta = y/x i = 1 * e^{i* pi/2} so when you do i^i it is 1^i * e^{i* pi/2 * i} = e^{-1 * pi/2} which is a real quantity
@@Linzz_1213 not necessarily. it is VERY EASY. to the point that you CAN understand it. maybe you don't have the intuition when it comes to complex numbwrs, which is okay, you could always learn. It seems like you're locking the concept behind some arbitrary wall you call "difficult math". No hate to anyone.
There’s even a proof that does not rely on Euler’s formula. Just take conjugate of i^i and manipulate to get i^i again. You could also first take the log before taking conjugate. Either way, only real numbers are unaffected by the complex conjugate.
I would ask your complex math teacher if this answer works for him.. If it does I would be concerned, since i is a ninety degree phase shift and a 90 degree phase shift of a ninety degree phase shift is meaningless in the complex plane. This should be solved in hypercomplex space and the question needs to have more defined boundary conditions or flakey math can be used to arrive at meaningless however clever answers.
Technically, we regard the complex logarithm as ill-defined. It just makes no sense as a concept that extends the real logarithm, for many issues. Coterminal angles are not the only issue. Contour integration makes this much more damning. That being said, the function Log(z) = ln(|z|) + atan2[Im(z), Re(z)]·i can be a useful one to define in some limited contexts, and since it is mildly analogous to the logarithm, the notation has stuck on, despite it being somewhat misleading. However, I think it would be healthier to avoid thinking of the above function as an actual logarithm. Superficially, it has some of the same properties, but it causes more conceptual problems than it solves. Anyway, if you interpret the notation ζ^ψ specifically as referring to exp[ψ·Log(ζ)], then it makes perfect sense to think of i^i as being equal to exp(-π/2). However, this interpretation of the notation is inconsistent with how exponents are typically treated. For example, it makes powers with base 0 undefined, and it also changes the odd fractional powers. Keeping in mind this inconsistency is something that many people need to be reminded of.
@@anshumanagrawal346 It makes sense only in a limited context. To be frank, even just raising real numbers to real numbers is already problematic. For example, a^b is treated as exp[b·ln(a)], but this is inconsistent with letting (-1)^2 = 1. However, there are many things that work with the logarithm for real numbers that make it a useful concept that do not translate for complex numbers. For one, the logarithm is a very important function in asymptotic analysis, and also very important in analytic number theory. Also, the natural logarithm can serve as an antiderivative of the reciprocal function. But for the complex numbers, the logarithm is not an antiderivative of the reciprocal function: the reciprocal function has no antiderivatives. And while the logarithm does not have the real numbers as the domain, only a proper subset instead, it at least is well-defined on that domain. In this sense, extending the logarithm to the complex numbers, rather fixing problems, only creates more. ln(0) is still undefined, but now ln is not even well-defined otherwise. Complex analysis has more power and theoretical elegance than real analysis, but it is not as though no sacrifices are being made when we work with complex numbers. I consider the concept of the logarithm to be one such sacrifice. In fact, the idea of restricting functions to some subset of the complex numbers for injectivity is less natural than it is with real numbers. Branch cuts are an extremely important idea in other areas of mathematics, but as far as functional analysis and complex arithmetic are concerned, it is very artificial. I am not saying the principal logarithm should never be used. I am saying the idea is nowhere near as natural an extension as many people think it is. And so, talking about it as if it was can be misleading at best.
@@dpage446 yeah except in this case we had numbers that were so cherished among mathematicians that they specially decided to dress them up with fancy alphabetical outfits i, e, π
I agree, if you only want to know what i^i is. I do think that Bri wanted to also show what ln(x+iy) would be, which makes this approach give the viewer a bit more knowledge than just the answer to the question in the thumbnail.
Wouldn't it be easier (and maybe less insightful) to substitute i by its polar form, exp(-i*pi/2)? Then you'll quickly find i^i = exp(i*-i*pi/2)=exp(pi/2) Edit: I missed a minus sign! Thanks for noticing!
I so FRICKEN LOVE THIS!!!! CRAZY AWESOME!!! I'm being serious here as I LOVE math and I have a huge math library full of math books in my home with four bookcases full of these books not counting an additional bookcase for my books on math history!!😁
Me too My mind is blown Being a 9th grader, Loving Math... not studied this but somehow was able to understand this thing...Its just Amazing I have made my own math book of math tricks... hopefully it will be published very soon and will be distributed to all math students 😀
@@demircan_demirbag I'm not sure what you mean. I have four bookcases full of general/popular math books and math textbooks and one bookcase for math history.
A real number (-1) raised to the power of a real number (0.5) gives an imaginary number (i) and raising an imaginary number (i) to the power of an imaginary number (i) gives a real number. Wow!
You did it correctly. I have seen other videos this subject and they skipped Ln(i) portion, even though they got the same answer, but wrong reasoning or technique.👌
I’m confused at 1:38. Sorry if this is a stupid question, im only in seventh grade. If r and e are multiplying, why is it ln(r) + ln(e^iθ), and not ln(r)*ln(e^iθ)?
Why so complicated? i^i=e^(i*i*Pi/2) directly using the Euler formula with i*i=-1 and then you get the real value i^i=e^(-Pi/2), so I don't understand this way of solving!
He did not apply logarithm to i but instead use Euler's identity to decompose i into polar form having arguments of r=1 and theta=pi/2. Both 1 and pi/2 are positive numbers so you can use the logarithmic function on both arguments.
Well, using Eulers Identify (e^ipi = -1) you can take the natural log of -1, as ln(-1) would be i*pi, you can use this to take the natural log of powers and multiples of i and negative numbers, and the natural log of -i (which also happens to be 1/i and i^3) as (3i*pi)/2. The natural log of i can be derived using properties of exponents and logs, as log (a^r) = r log (a), so ln (-1)^1/2 (which is square root of negative one), being 1/2 *ln (-1) which was stated to be i*pi earlier, giving up 1/2 * i*pi, giving up (i*pi)/2
I wonder what the significance of all possible rotations are for this number. Like why does it create exponential decay or growth of all things for when k is not 0 (usual convention).
Hey @brithemathguy, quick question, how do you find the derivative of y=f(x) while x=g(y). In this case, these functions are not inverses nor involutions. It looks like this function shoots up to infinity, can you help me with this?
@@Blaqjaqshellaq no, it is not. A lot of people are thinking that g(x) will be an inverse or an involution but it is none. The easiest example of what I'm saying is say f(x)= 5y^2+7y+1 and then let g(y)=3x^3+6. These functions are two independent algebraic functions, not inverses or involutions yet related to each other
@@Glitch-cp8wz If those two functions really depend on each other, then the resultant graph is just two points: (-1.32, -0.86) and (-1.30, -0.52). These are the points where the individual graphs intersect. The graph of the function is just a plot of two points and is thus not differentiable. You are effectively trying to differentiate a discrete list
@@AuroraNora3 oh okay. Also how did you get these points? I'm curious about that (I'm just a high school student doing some calculus, so i don't know anything much about it yet)
People need to realize that ln of x and ln of z are two different things. In the argand diagram complex functions are SURFACES. These are multi valued functions.
Pretty obvious. Power of i means a 90deg counterclockwise rotation,,,, Rotation is thus from the imaginary direction pi/2 (90 deg) to the real one pi (180 deg)... True for every 360 deg additional rotation.
Tbh i have an easier proof ,so it goes like this : i=√-1 Let's use eulers identity to take -1 to the complex world -1=e^(iπ) So i=√-1=e^(iπ/2) We are looking for i^i so lets's use other form of i we have just found (e^(iπ/2))^i In this expression we can multiply the exponent and we will have i × i which is -1 so it becomes e^(-π/2)which is also √(1/e^π) Which is a real number
My favorite one is exp(ℼi) = 1, when we redefine correctly ℼ = C/r. People should be using 'exp' more and more: For example, ln(xᵃ) = a ln(x) vs exp(ax) = exp(x)ᵃ
we can prove this without polar coordinates too. i^i=e^ln(i^i)=e^(ilni) ln(i)=ln((-1)^(1/2))=(1/2)(ln(-1)) ln(-1)=(i*pi) because of eulers identity =>(1/2)(ln(-1))=(i*pi)/2 =>i^i=e^i(i*pi/2)=e^-pi/2
Btw, it is not the only solution. You just had a principal solution, but you should note that it is multivalued function. Let me show it. i = e^(iπ/2) = e^(i5π/2) = e^(i9π/2) = ... So you will have: i^i = e^(iπ/2)i = e^(-π/2) i^i = e^(i5π/2)i = e^(-5π/2) i^i = e^(i9π/2)i = e^(-9π/2) ...
Не факт, что так! Условие, что любое выражение равно числу "е" в степени логарифм натуральный от этого выражения - такое действует возможно на действительные числа только, а вот работает ли так же оно и на мнимые - не факт
There exists a *much easier* way to find the *answer of i^i* , which *doesn't require us to find ln(i)* : We can write *i* as *0+1i* ==> *r* = sqrt(0^2+1^2) = *1* ==> *angle* = arctan(1/0) = arctan(undefined) = *π/2* Since *angle x* = angle x + 2πn (n belonging to the set of integers), and *a+bi = r*e^(i*angle)* , *i = e^[i*(π/2+2πn)]* => *i^i = e^[-(π/2+2πn)] Ans.* {multiplying the exponents to get i*i = -1 outside the parentheses} And we are *DONE* !
Nice, but I find it a bit unsatisfying. You just reach a number in decimal notation which kind of obscures the meaning of the number. But I guess you got me thinking on a different direction so I'll give you that
I have no idea what are you talking in video but I can solve it by e^πi =-1 Like this e^πi=i^2 e^π=i^(2/i) e^πi/2=i^1/I Since -1 is i^2 it caan be like this e^-π/2=i^i
> we could travel around the circle another 2pi and we would get a different result > when people say i to the u they usually mean this Bro Comon It doesn't matter how many times you go aroubd the circle i to the i is e to the minus half pi. You're not going to get a different result.
Or you could do it the other way, without log properties: x^y = e^(ln(x)y) that's the definition of e i^i = e^(ln(i)i) e^(i*pi) = -1 sqrt(e^(i*pi)) = sqrt(-1) e^((i*pi)/2) = i ln(i) = (i*pi)/2 i^i = e^(i(i*pi)/2) i^i = e^(-pi/2)
This may be an over simplification. i is a ninety degree phase shift. It being the square root of -1 is a symptom of that. If you just use the coordinate system of complex numbers you get 5pi/2 but that is not a 90 degree phase shift so the question is underbound and gives answers that may be nonsense. If you just use the complex plane then you have real numbers that are sampled at pi/2 but grow exponentially, well shrink exponentially using your math, so I think the answer is meaningless since what is a 90 degree phase shift of a 90 degree phase shift. If you use hyper complex numbers you would get ij =k for that and the answer would be e^-k(pi)/2 or i on the k axis. I think this answer is a little contrived.
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Hello!
i^i^... i times
really simple one, 1/i = -i
Zeta(i) lol (idk how dis works but yeah)
i-th root of i
I love the fact that a imaginary number just turned into a real number.
Yeah, it’s pretty hard to…imagine. 😎
@don'tfeel I’m curious, and as this question shows I am not very knowledgeable: how is the fact that i^i is equal to a real number related to hyperbolic geometry?
@don'tfeel Interesting; thanks for letting me know!
@@PunmasterSTP Joke aside, its actually not hard to imagine at all.
4^(1/3)
both of these numbers are rational
the output is irrational
but it's fine because both rational and irrational numbers are Real
similarly, both imaginary and real numbers are Complex!
@@wraithlordkoto Oh I quite agree; I just try to sneak in puns where I can 😎
I also think it’s interesting how one irrational number can be raised to the power of another irrational number, and end up being an integer!
Wow, something fun to watch that I don’t understand but appreciate the math!
Kinda easy ngl
@@mcsyllesen5183 do it then.
@@shaunthedcoaddict1656 look, every complex number z = x + iy can be represented as it's magnitude |z|(which is a real value), times it's "rotation" which is expressed as e^{i*theta}, where theta is the angle the line from 0,0 to the point z in the argand plane, or tan theta = y/x
i = 1 * e^{i* pi/2}
so when you do i^i
it is 1^i * e^{i* pi/2 * i}
= e^{-1 * pi/2}
which is a real quantity
@@Linzz_1213 not necessarily. it is VERY EASY. to the point that you CAN understand it. maybe you don't have the intuition when it comes to complex numbwrs, which is okay, you could always learn. It seems like you're locking the concept behind some arbitrary wall you call "difficult math". No hate to anyone.
same
you literally brought our imaginations to life( turned an imaginary number to a real number), thanks a lot
Thanks!
You bet!
There's a shorter (one line) solution:
e^(ipi) = -1 now raise left and right to the power i/2 and you get i^i = e^(-pi/2)
There’s even a proof that does not rely on Euler’s formula. Just take conjugate of i^i and manipulate to get i^i again. You could also first take the log before taking conjugate.
Either way, only real numbers are unaffected by the complex conjugate.
That's 2 lines
Multiplying the exponents may or may not apply to complex numbers
@@syed3344is there any proof confirming this identity or disproving it? Like it's troublesome to think that a proof is based off of thin ice
@@syed3344It always applies, (aᵇ)ᶜ=aᵇᶜ ∀a,b,c∈ℂ (a≠0). Just be careful not to change branches, x=y does not mean that xᵃ=yᵃ.
Kinda funny over the last weeks it seems like you’re uploading all the problems our complex analysis prof is handing to us haha
Great video!
I would ask your complex math teacher if this answer works for him.. If it does I would be concerned, since i is a ninety degree phase shift and a 90 degree phase shift of a ninety degree phase shift is meaningless in the complex plane. This should be solved in hypercomplex space and the question needs to have more defined boundary conditions or flakey math can be used to arrive at meaningless however clever answers.
@johnnychinstrap what a bitchy, unnecessary answer 🙄
We can do this simply putting value of i=e^(i*pi/2)
Then i^i=e^(i^2*pi/2)
So i^i=e^(-pi/2) {since i^2=-1}
Technically, we regard the complex logarithm as ill-defined. It just makes no sense as a concept that extends the real logarithm, for many issues. Coterminal angles are not the only issue. Contour integration makes this much more damning. That being said, the function Log(z) = ln(|z|) + atan2[Im(z), Re(z)]·i can be a useful one to define in some limited contexts, and since it is mildly analogous to the logarithm, the notation has stuck on, despite it being somewhat misleading. However, I think it would be healthier to avoid thinking of the above function as an actual logarithm. Superficially, it has some of the same properties, but it causes more conceptual problems than it solves.
Anyway, if you interpret the notation ζ^ψ specifically as referring to exp[ψ·Log(ζ)], then it makes perfect sense to think of i^i as being equal to exp(-π/2). However, this interpretation of the notation is inconsistent with how exponents are typically treated. For example, it makes powers with base 0 undefined, and it also changes the odd fractional powers. Keeping in mind this inconsistency is something that many people need to be reminded of.
So then raising a not purely real complex no. to an imaginary power doesn't make sense in general, right?
@@anshumanagrawal346 It makes sense only in a limited context. To be frank, even just raising real numbers to real numbers is already problematic. For example, a^b is treated as exp[b·ln(a)], but this is inconsistent with letting (-1)^2 = 1. However, there are many things that work with the logarithm for real numbers that make it a useful concept that do not translate for complex numbers. For one, the logarithm is a very important function in asymptotic analysis, and also very important in analytic number theory. Also, the natural logarithm can serve as an antiderivative of the reciprocal function. But for the complex numbers, the logarithm is not an antiderivative of the reciprocal function: the reciprocal function has no antiderivatives. And while the logarithm does not have the real numbers as the domain, only a proper subset instead, it at least is well-defined on that domain. In this sense, extending the logarithm to the complex numbers, rather fixing problems, only creates more. ln(0) is still undefined, but now ln is not even well-defined otherwise.
Complex analysis has more power and theoretical elegance than real analysis, but it is not as though no sacrifices are being made when we work with complex numbers. I consider the concept of the logarithm to be one such sacrifice. In fact, the idea of restricting functions to some subset of the complex numbers for injectivity is less natural than it is with real numbers. Branch cuts are an extremely important idea in other areas of mathematics, but as far as functional analysis and complex arithmetic are concerned, it is very artificial. I am not saying the principal logarithm should never be used. I am saying the idea is nowhere near as natural an extension as many people think it is. And so, talking about it as if it was can be misleading at best.
Found the fun-ruiner
Logarithms are multifunctions.
you dont need the log tho. u can jus substitute using eulers identity. e^iπ=-1, e^iπ/2=i then i^i=(e^iπ/2)^i=e^(i^2)π/2=e^-π/2. no logs
The fact that he is doing maths but I cant see a single number the whole video scares me
Math without numbers>math with numbers
E is a constant
@@lakshay-musicalscientist2144 i is also a number, just an imaginary one
I know that but i assumed that the guy meant a numerical coefficient of an expression
@@dpage446 yeah except in this case we had numbers that were so cherished among mathematicians that they specially decided to dress them up with fancy alphabetical outfits i, e, π
I think it's much easier this way:
i^i = (e^(iπ/2))^i = e^(π/2 * i^2) = e^(-π/2)
I agree, if you only want to know what i^i is. I do think that Bri wanted to also show what ln(x+iy) would be, which makes this approach give the viewer a bit more knowledge than just the answer to the question in the thumbnail.
The title
Was so accurate
I yelled wtf when I realized what was going on and my mom came to see if everything's fine lol
amazing video
1:49 that r is still stuck inside of that ln
Wouldn't it be easier (and maybe less insightful) to substitute i by its polar form, exp(-i*pi/2)?
Then you'll quickly find i^i = exp(i*-i*pi/2)=exp(pi/2)
Edit: I missed a minus sign! Thanks for noticing!
That's what i was thinking when i saw the thumbnail
U are missing a minus on exp(pi/2)
Yes, this is correct and a much easier way to solve the problem (assuming you put in the aforementioned minus sign)
This is wrong!!
That polar form is jus tone of infinite answers, the real substitution is i=exp(i*(pi/2+2*pi*k)), where k is an integer.
I so FRICKEN LOVE THIS!!!! CRAZY AWESOME!!! I'm being serious here as I LOVE math and I have a huge math library full of math books in my home with four bookcases full of these books not counting an additional bookcase for my books on math history!!😁
🤓
Me too
My mind is blown
Being a 9th grader, Loving Math... not studied this but somehow was able to understand this thing...Its just Amazing
I have made my own math book of math tricks... hopefully it will be published very soon and will be distributed to all math students 😀
@@IS-py3dk OH MY! Can you please publish it?
@Pine Delgado and you have another bookcase which contains the books which describe the places of your books. Right?
@@demircan_demirbag I'm not sure what you mean. I have four bookcases full of general/popular math books and math textbooks and one bookcase for math history.
A real number (-1) raised to the power of a real number (0.5) gives an imaginary number (i) and raising an imaginary number (i) to the power of an imaginary number (i) gives a real number. Wow!
Since i is defined as the square root of a real number, it's interesting, but also obvious at least in retrospect.
One thing I like to do is apply Euler's formula to i ln(i) to get cos(ln(i)) + i sin(ln(i)). Not much point to doing this but it's kinda fun.
I personally just substitute the i in the base for e^(i*pi/2), raise that to the ith power, replace i*i with -1, and get e^(-pi/2)
I didn't even bother taking the natural log.
I knew z=re^i*theta
hence, i = e^(i*pi/2)
thus i^i = (e^(i*pi/2))^i
=> i^i = e^(i*i*pi/2)
=> i^i = e^(-pi/2)
That title is perfect lmfao
"i to the i, it's about a fifth" - Matt Parker of the legendary Parker Square
I don't exactly understand the step he takes at 2:34. How does ln(i) = i(pi)/2 turn to e^i *i(pi)/2?
He substitutes ln(i) as i(pi)/2 in the expression e^i * ln(i)
the inverse of ln is e, therefore raising e to the ln(i) cancels the ln and the e
z-->i^z is a multivalued function. You can't attribute a unique value to i^i.
I'm glad we see eye to eye on this one.
You did it correctly. I have seen other videos this subject and they skipped Ln(i) portion, even though they got the same answer, but wrong reasoning or technique.👌
I’m confused at 1:38. Sorry if this is a stupid question, im only in seventh grade.
If r and e are multiplying, why is it ln(r) + ln(e^iθ), and not ln(r)*ln(e^iθ)?
That's just how ln() works. he even put the line in yellow to tell you :w
Why we can let ln(i^i) eq i*ln(i)?
ln(x^a)=a ln(x) for x, a in R.. but i in Z.
Why so complicated? i^i=e^(i*i*Pi/2) directly using the Euler formula with i*i=-1 and then you get the real value i^i=e^(-Pi/2), so I don't understand this way of solving!
You can apply logarithm only to the argument that is positive real number. Here we are not sure, i^i is positive or not.
He did not apply logarithm to i but instead use Euler's identity to decompose i into polar form having arguments of r=1 and theta=pi/2. Both 1 and pi/2 are positive numbers so you can use the logarithmic function on both arguments.
Well, using Eulers Identify (e^ipi = -1) you can take the natural log of -1, as ln(-1) would be i*pi, you can use this to take the natural log of powers and multiples of i and negative numbers, and the natural log of -i (which also happens to be 1/i and i^3) as (3i*pi)/2. The natural log of i can be derived using properties of exponents and logs, as log (a^r) = r log (a), so ln (-1)^1/2 (which is square root of negative one), being 1/2 *ln (-1) which was stated to be i*pi earlier, giving up 1/2 * i*pi, giving up (i*pi)/2
@@thegddevil5664 nuff said.
I wonder what the significance of all possible rotations are for this number. Like why does it create exponential decay or growth of all things for when k is not 0 (usual convention).
possible to cover the principal value in complex? this might be the key to unlock summing to infinity
Hey @brithemathguy, quick question, how do you find the derivative of y=f(x) while x=g(y). In this case, these functions are not inverses nor involutions. It looks like this function shoots up to infinity, can you help me with this?
@@cameronbigley7483 i don't think in any of these functions y and x are separable in any way because each others are arguments of each other
f'(x) will be the inverse of g'(y)
@@Blaqjaqshellaq no, it is not. A lot of people are thinking that g(x) will be an inverse or an involution but it is none. The easiest example of what I'm saying is say f(x)= 5y^2+7y+1 and then let g(y)=3x^3+6. These functions are two independent algebraic functions, not inverses or involutions yet related to each other
@@Glitch-cp8wz If those two functions really depend on each other, then the resultant graph is just two points:
(-1.32, -0.86) and (-1.30, -0.52). These are the points where the individual graphs intersect.
The graph of the function is just a plot of two points and is thus not differentiable. You are effectively trying to differentiate a discrete list
@@AuroraNora3 oh okay. Also how did you get these points? I'm curious about that (I'm just a high school student doing some calculus, so i don't know anything much about it yet)
My calculator said Math error😂 but for a good reason that we can't specifically define it since there are infinite solutions.
using i in the exponential form would have been way quicker
Just replace i by e^(i * pi/2)
The fact that (a^b)^c = a^bc if a is a positive number and b and c are real numbers and you can’t use it for complex numbers
My brain ceases to function when any kind of algebra is involved.
But this still seems like a great video 👍
e^(i*π) = -1 | ln(...)
i*π = ln(-1) = ln(i^2) = 2*ln(i) | * 1/2
i*π/2 = ln(i)
branch cuts???
You always win my heart !
People need to realize that ln of x and ln of z are two different things. In the argand diagram complex functions are SURFACES. These are multi valued functions.
Pretty obvious. Power of i means a 90deg counterclockwise rotation,,,, Rotation is thus from the imaginary direction pi/2 (90 deg) to the real one pi (180 deg)... True for every 360 deg additional rotation.
You have to define i^i ;
i^n is i*....*i n times.
But what is i^i.....?
how did he get re^itheta at 1:25
Does anyone know the pun in the title
Very nice, Euler would be happy.
the fact i found this just by seeing the thumbnail makes me proud of myself
Well done
You can start solution based on the fact that e^iπ = -1 and solve the problem.
I knew the answer before hand, and when I saw the title and thumbnail, I couldn't stop laughing, this can't be real!
Once you put wrote i^i as e^ln(i^i) it was obvious to me, it's awesome how cool yet simple this is
Tbh i have an easier proof ,so it goes like this :
i=√-1
Let's use eulers identity to take -1 to the complex world
-1=e^(iπ)
So
i=√-1=e^(iπ/2)
We are looking for i^i so lets's use other form of i we have just found
(e^(iπ/2))^i
In this expression we can multiply the exponent and we will have i × i which is -1 so it becomes
e^(-π/2)which is also √(1/e^π)
Which is a real number
You can apply Euler's formula in the beginn and get this done much more quickly
My favorite one is exp(ℼi) = 1, when we redefine correctly ℼ = C/r.
People should be using 'exp' more and more:
For example, ln(xᵃ) = a ln(x) vs exp(ax) = exp(x)ᵃ
bro.. that's not the π symbol..
Write i in exponential form, then raise to power of i.
we can prove this without polar coordinates too. i^i=e^ln(i^i)=e^(ilni)
ln(i)=ln((-1)^(1/2))=(1/2)(ln(-1))
ln(-1)=(i*pi) because of eulers identity
=>(1/2)(ln(-1))=(i*pi)/2
=>i^i=e^i(i*pi/2)=e^-pi/2
Btw, it is not the only solution. You just had a principal solution, but you should note that it is multivalued function. Let me show it.
i = e^(iπ/2) = e^(i5π/2) = e^(i9π/2) = ...
So you will have:
i^i = e^(iπ/2)i = e^(-π/2)
i^i = e^(i5π/2)i = e^(-5π/2)
i^i = e^(i9π/2)i = e^(-9π/2)
...
as if it wasn't already stated in the video...
thought you were gonna use e^(ix) when x=ln(i)
Не факт, что так! Условие, что любое выражение равно числу "е" в степени логарифм натуральный от этого выражения - такое действует возможно на действительные числа только, а вот работает ли так же оно и на мнимые - не факт
Well, i understand this but how can we get the idea to do this??
I see what you did there 😂😂😂
can you please make a math book list recommendation?
But can we apply ln to an complex number
Very educational.Thank you❤
this is beautiful, I love your channel!
I just finished imaginary numbers a few weeks ago thinking I'll understand this video. NEVERMIND
Didn't BPRP do this like years ago?
Can't you just do it like this:
e^ipi=-1
then sqrt both sides
sqrt(e^ipi)=i
then raise both sides to the power of i
(sqrt(e^ipi)^i)=i^i
=e^-pi/2
That was just awesome 🔥
Aren't roots the opposite of exponents. For example when something is squared you will take the square root to undo it
yes but no
Well, this is complex numbers made even more complex.
_Complexer Numbers_
Imagine being so imaginative that u turned Real
e^(-п/2), как главное значение.
The funny part about this video is the title, because the solution IS actually real.
There exists a *much easier* way to find the *answer of i^i* , which *doesn't require us to find ln(i)* :
We can write *i* as *0+1i*
==> *r* = sqrt(0^2+1^2) = *1*
==> *angle* = arctan(1/0) = arctan(undefined) = *π/2*
Since *angle x* = angle x + 2πn (n belonging to the set of integers), and *a+bi = r*e^(i*angle)* ,
*i = e^[i*(π/2+2πn)]*
=> *i^i = e^[-(π/2+2πn)] Ans.*
{multiplying the exponents to get i*i = -1 outside the parentheses}
And we are *DONE* !
I just learnt so much and nothing at the same time.
this is incorrect i^i is not defined because it can be e^npi/2 where n any integer. ln(i) is not defined. define ln(i) first in a consistent way.
Therapist: i^i isnt real, it cant hurt you
i^i:
But what it geometrically means ?
When i saw the thumbnail, i thought 'Not real, this is complex'
But damn is it actually real
Nice, but I find it a bit unsatisfying. You just reach a number in decimal notation which kind of obscures the meaning of the number. But I guess you got me thinking on a different direction so I'll give you that
but he showed few seconds before that its meaning was e^-(π/2)
what
I have no idea what are you talking in video but I can solve it by e^πi =-1
Like this
e^πi=i^2
e^π=i^(2/i)
e^πi/2=i^1/I
Since -1 is i^2 it caan be like this
e^-π/2=i^i
Much faster and more elegant.
@@Lolwutdesu9000 thank you !
Jeje 😄
perfectly explained.
> we could travel around the circle another 2pi and we would get a different result
> when people say i to the u they usually mean this
Bro
Comon
It doesn't matter how many times you go aroubd the circle i to the i is e to the minus half pi. You're not going to get a different result.
1 to the i on the other hand does have multiple values,
The way I solved it:
i = e^(i*pi/2)
i^i = e^(i*pi/2)^i
= e^(i^2 * pi/2)
= e^(-pi/2)
Or you could do it the other way, without log properties:
x^y = e^(ln(x)y) that's the definition of e
i^i = e^(ln(i)i)
e^(i*pi) = -1
sqrt(e^(i*pi)) = sqrt(-1)
e^((i*pi)/2) = i
ln(i) = (i*pi)/2
i^i = e^(i(i*pi)/2)
i^i = e^(-pi/2)
no because ln is not defined on complex numbers. define the ln(i)
@@tokajileo5928 ln(z) = ln(r)+ti
Where z = r*e^ti
How do you divide by 0? Did you make a video about it yet?Try Physics sometime , thanks 😊.
He did make a video on it.
damn thats cool but what nubmer comes after 8 again ?
Caption: 'this just can't be real'
Me: wrong on a few levels, there...
This may be an over simplification. i is a ninety degree phase shift. It being the square root of -1 is a symptom of that. If you just use the coordinate system of complex numbers you get 5pi/2 but that is not a 90 degree phase shift so the question is underbound and gives answers that may be nonsense. If you just use the complex plane then you have real numbers that are sampled at pi/2 but grow exponentially, well shrink exponentially using your math, so I think the answer is meaningless since what is a 90 degree phase shift of a 90 degree phase shift. If you use hyper complex numbers you would get ij =k for that and the answer would be e^-k(pi)/2 or i on the k axis. I think this answer is a little contrived.
Natural log is equal to pi/2
“This just can’t be real” no pun intended
How am i watching this not understanding anything but still finding it interesting somehow
What grade is this?
So, in a set of number equal to the complex set i^i is undefined.
this just can’t be real
i^i: i’m gonna end this man’s whole career
i^i is not only real look sick and it contains e and π like what else do you want from a number 😆.
imagination with the power of imagination becomes real?!? i need to imagine stronger.
i^i = sinpi/15
What's the cube root of -1
irrational complex number
@@Ostup_Burtik no, it's -1
@@JustAnIdiot6969 wrong
@@Ostup_Burtik read the question, it asks cube root, not the square root.
@@JustAnIdiot6969 yes, and ³√-1 is not -1.
The i root of that number is equal to i