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- Опубликовано: 28 сен 2024
- We will find both the square roots of i, i.e. sqrt(i). We will first write sqrt(i) as a complex number a+bi and then square both sides. Then we will solve for a and b by setting a system of equations! This is the algebra way to find the square root of the imaginary unit i.
Check out these related videos:
polar way: • sqrt(i) in polar form
sqrt(i+sqrt(i+sqrt(i+...))): • sqrt(i+sqrt(i+sqrt(i+....
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sqrt(a+bi)=?
Answer here: ruclips.net/video/CeVdh5LH908/видео.html
👍
White Chalk Red Chalk, nice 😊
Steve:
0:00 As we all know, √(-1) = 𝒊
Also Steve:
8:50 √𝒊 = ±(1 + 𝒊)/√2 - *two answers*
That's inconsistent!
@@allozovsky it's not inconsistent. There's one in Q I and another in Q III separated by p u/2 rotation Much like using an inverse trig function to return an answer in a limited domain where the actual solution may be outside of that domain. Recall that -1 is a real number, I is imaginary, and thr square roots are complex. Complex numbers don't behave quite the same as real and purely imaginary numbers.
@@onradioactivewaves If √𝒊 = ±(1 + 𝒊)/√2 gives two complex square roots, then √(−1) also should return two complex square roots, that is √(−1) = ±𝒊, isn't it? Otherwise it is inconsistent. That's pretty strange that Steve often evaluates in his videos multivalued complex functions alright, but at the same time uses only single 𝒊 for the square root of −1.
I happened to watch this on my break from studying before my leaving cert maths exam. Square root of a complex number was on the exam and I got the right answer using this method. What a crazy lucky coincidence.
Congrats man
Yeah I remember the exam. I used a different method tho.
@@artemis_furrson What was it? 🧐
@Mosinlogan Being interested in what you study is a blessing many students would strive to have.
@@Someone-wj1lf there's no way people are like that? Lol I didn't know that, that's strange
I don't know why people keep complaining about this guy's solution to the problem, and why they offer geometric proofs instead. I really like this guy's answer because it uses only the simplest arithmetic/algebra and the simplest definition of a complex number: a + bi. Also, he did it in an incredibly concise manner. (My only complaint was at the end where he could have used a² - b² = (a+b)(a-b) = 0 --> a = ±b. But, typing this out, I suddenly realise how clever he is that even *factorization* doesn't need to be used in his answer.) This answer is teachable on someone's very first lesson on complex numbers; even the average 15-year-old will comprehend it very well, and he's earned my amazement.
Billy Ma-gusta thank you Billy. I think people just got too excited. It's like once they see derivatives, they want to show algebra students to take the derivative to find the vertex of a parabola.
Billy Ma-gusta You want concise?
Polar form: R*exp(iθ) = sqrt(i)
Square: R^2*exp(2iθ) = i = exp(i(π/2 + 2πn)), n integer
Identify: R=1, θ=π/4+nπ
Done.
Or if you prefer rectangular form the unique representations are: ±(1+i)/sqrt(2)
Tl;dr: Rectangular is great for addition and subtraction, polar for multiplication, powers and roots. Right tool for the job.
My point: Your working is as opaque as it gets. For all the average high schooler or RUclipsr can see, you're just writing a whole bunch of Greek because nothing is explained. The polar form of complex numbers isn't explained at all. The polar form of √i specifically also isn't explained, neither is i. Converting between the polar form and the rectangular form also isn't explained. You cannot call this concise when you leave so much unexplained; someone who doesn't know the required background knowledge would call this gibberish. (Like how I call bullshit when I am told "using string theory, we can show that 11 physical dimensions exist". Just an example, don't digress pls.)
And the greater point: the polar form does not NEED to be explained in order to prove this result, as bprp has shown. Simpler is better in maths, and a short proof is not necessarily a simple proof.
I would happily agree that I assume too much if I for some video on an
integral complain about someone not using residuals and Cauchy's
integral formula to solve an integral on the real line since it would be
trivial that way. But assuming knowledge of the polar form of a complex number, which is as fundamental as the rectangular form, is not a stretch imo.
As for the specific polar forms. I never use the polar form of sqrt(i), only the polar form of i, to keep things simpler. I could happily have shown a justification of the polar form of i, but it's a bit clunky without being able to show any figures. Basically you have an angle of π/2 to i and any 2π increment of that still leaves you at i. If I had shown this on a blackboard the justification is explained in a few seconds.
The conversion to rectangular form is really not part of my solution, I'm happy to stay in polar form.
My point is that this result is trivial in the polar world. It also gives some really nice insights about how the root behaves in general, which is totally lost in rectangular form. Using polar form the result also trivially generalizes to any real power of i.
Just like addition/subtraction is trivial in rectangular but a mess in polar form. Sure you could technically use the strategy as shown in this video to describe the 5th or 50th root of i as well, but it would be very cumbersome and very brute force.
you actually don't need factorization to get that a = b.
1. a² - b² = 0 -> a² = b², therefore a and b have the same magnitude.
2. 2ab = 1, ab = 1/2, therefore a and b have the same sign (if the product of 2 real numbers is positive, then 2 real numbers have the same sign)
3. a = b, because a and b have the same magnitude (1) and the same sign (2) (definition of equality for real numbers)
4.a*a = 1/2 = a², from (3) and (2)
5. a = ± sqrt(1/2), sqrt both sides of (4)
6. b = ± sqrt(1/2), from (3)
I actually prefer this method of solving the system of equations over the video because it takes less steps and is more intuitive. the videos method involves a lot of seemingly arbitrary moving of symbols around while each of my 6 steps have much clearer purpose.
I, as a non native english speaker, watched your video at 2x speed. Got everything you said. Keep it up.
You should be an English teacher
Cough cough the pinned comment cough cough
me too, I guess as a non english native speaker you are used to a bigger variety of accents
OMG EXACTLY ME!!!!!!!!!!
Me too. I think improving the accent would be nice still, though.
What I miss when I zone out for 30 seconds in class:
Underrated comment
ADHD moment. Felt it.
@@kdog3908what is ADHD!?
@@swetkataria5282 Attention Deficit Hyperactivity Disorder. More commonly called ADD in the US, I think.
There ya go psychologists; the root of imagination.
This is so underrated
This deserves more likes...
Bro what 30 likes 😂
The blackboard was designed by a guy named Hilbert.
Woo
Plot twist: He has endless layers of boards.
Hahhaa
Truuue
blackboardwhiteboard
😂😂😂
@BLVGaming / Y1000 Couldn't not agree
Me: trying to go to sleep
RUclips: BuT wHaTs ThE sQuArE rOoT oF i???
-1
@@laerciocivali no
lol
@@laerciocivali thats i squared
Same here XD
You can also convert to e^(i*π/2). Then sqrt(e^(i*π/2))=e^(i*π/4). Then convert back to get 1/sqrt(2)+i/sqrt(2)
And losing second solution )
You can get the second solution by generally writing sqrt(i)=exp(i*(pi/2+k*pi)) for k any integer
@@CorvusSapien It's not a solution. You write just answer. Initial post suppose to write i, but not sqrt(i) in exponential form. And than use powering properties.
One could write
i = exp(i*π/2 + 2πk)
i**0.5 = exp(i*π/4 + πk)
But why we didn't lose something else in this solution?
Using powering properties is totally wrong way of thinking in this case.
@@at_one doing square root gives 2 solutions
sqrt(i)= ±sqrt(e^πi/2)= ±e^πi/4 = ±(cos(π/4)+isin(π/4))
giving the 2 solutions: sqrt(i)=1/sqrt(2) + i/sqrt(2) or sqrt(i)=-1/sqrt(2) -i/sqrt(2)
@@peted2783 this is not a question. The question is about using powering properties.
My opinion is that your's approach is wrong. On this case it gives correct answer, but in general case not.
One must use this formula while square rooting:
en.m.wikipedia.org/wiki/De_Moivre%27s_formula
But you shouldn't use this formula:
(e^z)^0.5 ≠ e^(0.5z)
to find all roots.
Initial post is about powering. And my comment is about it. I know how to square rooting in complex field 😂
There is an other easy way :
We have i=e^(i(π/2))
So √i =[e^(i(π/2))]^((1/2))
So √i=e^(i(π/4))=√2/2 +i √2/2
It means that √i = √2/2 +i √2/2
Brilliant 👍
Damn bro nice flex lol
Aka 1+i
i don't understand
@@ericw2391 ok l will check it.Thanks+
I appreciate the depth of this explanation, rather than memorizing forms, for the sake of speed and ease, you showed me understanding. I appreciate knowing why, over mechanical speed.
You'll get mechanical speed with practice. Idiots just don't know to practice.
@@woophereigo9755 Smart guy over here. Shut up.
@@manperson6234 Bunch of morons. Get good.
@@woophereigo9755 yikes bud 😬
@@woophereigo9755 One thing is be fast while doing your own calculations, another is following another one being fast doing his calculations
6:05 when he got the answer he started moving closer to the speed of light
Lmao underrated
When you beat a boss in Kingdom Hearts
@@francescolorenzelli8912 didnt think id find a kingdom hearts reference here ..... i still dont get the joke tho
edit: Ahh i see the vision is blury...
lmaoo
I thought the same thing haha
Wanna know what's behind my board? It's another board!
bruh this teacher is so prepared. he uses 2 different colors of chalk to distinguish between terms and grouping symbols. Good Job!
Wanna know what's behind the second board? A third board!
@@_carrbgamingjr yep
I'm board.
@@stevens5541 ok
I litterally read it as "Squ(i)rt"
I'll remember that forever
ㅋㅋㅋㅋㅋㅋㅋㅋ
@@o_5553 same
Time to stop waxing your carrot
Me too
Only thing confusing was saying 1/(2*(1/√2)) = 1/√2
When I looked I instead got √2/2 but if you multiply by √2/√2 you get 2/2√2 which gives you 1/√2 so you were right but that part was the only thing I found to be unclear
glad you commented because that tripped me up as well
@@TheWannaramble 2/sqrt(2) should be multiplied with sqrt(2)/sqrt(2) (which changes nothing since it's just 1).
2*sqrt(2)/sqrt(2)*sqrt(2)
We know that square root of n multiplied with itself gives us n so:
2*sqrt(2)/2
Both sides divided by 2
Sqrt(2)
@@someoneunimportant3064 very clear, thanks
@@TheWannaramble you are very welcome, glad it helped
@@TheWannaramble that is what I thought
Another way can be e^(i*π/2)= i for r=1, square root both sides and it will be e^(i*π/4)=√i, which will give √i= (1+i)√2
that only gives the positive sides of things because you're doing it in polar form which makes the angle be divided by two, and it was positive (90 cuz e^(pi*i) is on the Y axis)
meaning it has an angle of 45 now, which is only the positive quarter, meaning you have to draw a 45 degree line, or Y = X line, and take the answers that sit on the unit circle
which are two
45 degrees, and the +180 degrees from that, 215. meaning you'll get the one in the negative quarter
whitechalkredchalk
Lol
**He lifts up the black board**
Me - what the hell is thisss?
*lifts second blackboard revealing a third*
“Oh my god!”
@@SuperUghe yeah I guess the future is here!
What? This is pretty common xD
@@SuperUghe It's blackboards all the way down.
No it is -green board-
Ok fine but, Why is he holding a grenade in his hands
that's a microphone
@@kayjaad3349 $Thanks, I did not know that_
@@kayjaad3349 there's something called sarcasm sis......
@AFancySpoon you'll only get attention if you comment on the board that this guy is using XD
*Bitter truth tho*
His parents made him hold it, it'll go off the moment he makes a mistake.
My professor taught me something valuable when writing my master’s: Never start an argument/discussion/presentation with ‘as we all know’; you never know who doesn’t know, and thus risk pushing away potentially interested readers.
I completely agree with your professor about that.
I think if youre a math channel that posts exclusively calculus content its okay to assume your viewers know the definition of i
@@callumross6290 What if you are interested in math but never learnt calculus, and someone recommended this channel as it’s really good content?
Make sure you don't focus too much on improving your accent first of all. I could comprehend it just fine. Chinese pronunciation works just fine on English language as long as you have practiced. More good videos please.
you are studying math you dont even know how to speak english to understand this. math is universa, i didnt even listen the audio to understand everything i jus tskipped it
@TheMasterfulcreator I see what you did there.
Nice I like what u did
Math is its own language
My calc class is taught by a Romanian woman, half the class is Chinese. Communication is not a problem. Math is the universal language, numbers unite us all.
My man’s too drippy for us, wearing supreme and teaching maths
😆
When the number is imaginary!
Ù⁸⁸
He’s on another level
Root_4 (-1)
I don't understand how this can be SO FUKING PERFECT
Bruno Amezcua, because mathematics is a series of quantitative tautologies, where each system builds off the previous system.
@@stumpfightskills571 Well said!!⚡🔥
He almost made a writing mistake at the end (= instead of or) but fixed it immediately...
You can also think about it in polar form. i is on the unit circle, so it's roots are also on the unit circle. The argument for the principal value must be π/4 (½ the argument of i). So if you have your unit circle memorized the principal root is clearly sqrt(2)/2 + i*sqrt(2)/2. The other root is opposite the principal root at -sqrt(2)/2 -i*sqrt(2)/2
Damn Its so much simpler
That's what I was think. You are rotating half way towards the imaginary number line from the real number line. That would be pi/4 rotation. Then figure out your polar coordinates and trig.
x = √i
x² = i
x⁴ = i² = -1
x⁴ + 1 = 0
x⁴ + 2x² + 1 = 0 + 2x²
(x²)² + 2x² + 1 = 2x²
(x² + 1)² = 2x²
x² + 1 = x √2
x² - x√2 = -1
x² - 2(x)(√2 / 2) = -1
x² - 2(x)(1/√2) + ½ = -1 + ½
x² - 2(x)(1/√2) + (1/√2)² = -½
(x - 1/√2)² = -½
x - 1/√2 = ±√(-½) = ± √(-1) / √2
x = 1/√2 ± i/√2
√i = 1/√2 (1 ± i)
I also solved the problem that way.
Yep, that's the way I learned it.
Very understandable even as a non native speaker
thanks!!
I agree. Your work is very good. Dont worry about trolls who complain and then fight.
*egg*
@@YellowToad egg
Es verdad mientras voy en un bus lo miro, por el alto volumen de bus no puedo oír el vídeo, soy hispanohablante, aún así se entiende todo.
Another way to see it is that multiplying by i makes the complex number rotate around origo by 90° (pi/2). Multiplying by i^(1/2) instead rotates 45° (pi/4). So, for example, 1 × i^(3/2) = -1/sqrt(2) + i/sqrt(2) since that is where a rotation of 135° from 0° takes us.
Thank you for sharing this.
8:54
10 people got *TRIGGERED* because he didn't rationalize the denominator
IAAGO ARIEL SCHWOELK LOBO lol!!!!
IAAGO ARIEL SCHWOELK LOBO No one rationalizes denominators in 2017 - that's what people did back in the day when there were no calculators and you'd prefer dividing the memorized decimal form of the square root by the rationalized denominator.
Another thing is that rationalizing the denominator often hides the geometric connections between quantities. It's a bit harder to see that `√2/2` is the inverse of `√2`, but it is obvious when you didn't rationalize it: `1/√2` (one over something is the inverse). It's even more hard to see it with some more complex expressions with radicals. That's why I usually leave it unrationalized, as an inverse, unless I really have to rationalize it.
Sci Twi I wonder what modern books have answer keys that use rationalised denominators..
Sci Twi Look; I've got mixed feelings for the conventions of rationalizing denominators or not. You are completely right about maintaining the instant recognition for inverses, but then you'd be compromising the recognition of like terms. For example, 3/sqrt5 does not look like it could be added to 2sqrt5/2, but after rationalizing, you can see clearly that 3sqrt5/5 _can_. Now, on the issue of having the same answers as the teachers do (and trying to overlook the insult to many great math teachers I've met that I'm sure you were not trying to offend), you can't really be opposed to unification of measures or answers -at least to some extent you have to accept it. Of course it makes your life easier to save extra moments on a test or whatever, but using a more real life example with more important implications, the SIU (International System of Units)'s purpose is to ease scientists' endeavors at "sciencing", if you will, by having set standards as to what units are official, what they measure, and how much of that something they measure. This, of course, may mean little to a mathematician's job, but if you can apply this same smooth interchange of information through the answers and numbers you represent, what you try to state will be better conveyed and understood by the audience to whom you present the information to. Anyway, I know I can't force someone to think the way I do, and you have to use the methods that you know are better for your learning (very similar to the π/τ argument), but thanks for reading to the end.
You went a really complicated way of solving these equations. In my head I did it like so
a^2 - b^2 = 0; a^2 = b^2; a = +-b
2ab = 1; we know a and b should be of the same sign so we'll say a = b and get
2a^2 = 1
a^2 = 1/2
a = sqrt(1/2)
yup.. same here
Exactly what I was thinking. But we would probably get less marks than him cause he did the longer method. 😂
Thats why so many people don't get into maths. Maths can be wathever you want, depending on how you enter in. Many examples show very simple solution or very complex ones for the same question. Which one do you prefer ?
Even simpler:
sqrt(i) = (e^(i pi/2))^(1/2) = e^(i pi/4) = cos(pi/4) + i sin(pi/4)
Even if signs are same It doesn't make a=b
All the people complaining he didn't use polar co-ordinates are completely missing the point. If you haven't already studied exactly why e^ix = i sin(x) + cos(x) then that would make this video completely pointless, the kind of people who want to know the answer to this problem most likely haven't come across that level of mathematics yet
Harry Stuart thanks!
Harry Stuart, well assuming you have a calculator for inverse Tan or a book full with tables of precalculated values for not carefully selected examples. And dividing is multiplying with the power of -1 defined as (a-bi)/(a^2+b^2) but you are free to learn such tables like some people like studying phone book numbers 😋
I agree that it doesn't look nice when you first come across it, but polar representation is one of the reasons complex numbers are so useful in the sciences
densch123 you'll wind up being thankful for the exponential form when you hit differential equations :P
or theoretical physics... or having to find real and imaginary roots of numbers... and complex functions and variables... lol ;P
This showed up on my recommended, and I could feel myself getting smarter throughout the video because of your amazing teaching style. You have earned yourself a new subscriber, so thank you
I may be the only one who liked you accent. And could you do a video on differential equations?
Pink Floyd is the Best Band of All Time. Hi there, thanks!! I do have diff eq videos here www.blackpenredpen.com/math/DiffEq.html
blackpenredpen Thanks!
I just realized, this answer has a magnitude of 1 on the complex number plane. If you just looked at a and b while disregarding the i, you could say that with Pythagorean’s Theorem and with a and b as the x and y coordinates, the hypotenuse is 1.
If you draw a unit circle with radius 1 on the complex plane (which touches points 1, -1, i, -i), then you can draw angles based on the points plotted on this unit circle. The angle of this answer with the real number 1 is 45 degrees. The angle between i and 1 is 90 degrees. The angle between -1 and 1 is 180 degrees. The angle between 1 and 1 is 360 degrees. This answer squared is i, i squared is -1, -1 squared is 1.
True, you can also solve multiplications using the complex plane: any two nubers mulitpilied will have an angle equal to the angle of the first and the angle of the second number summed up, and will have a distance from 0 equal to the distance of the first number multiplied by the distance of the second number
I’m french sorry if this isn’t very clear
x = √i
x² = i
x⁴ = i² = -1
x⁴ + 1 = 0
x⁴ + 2x² + 1 = 0 + 2x²
(x²)² + 2x² + 1 = 2x²
(x² + 1)² = 2x²
x² + 1 = x √2
x² - x√2 = -1
x² - 2(x)(√2 / 2) = -1
x² - 2(x)(1/√2) + ½ = -1 + ½
x² - 2(x)(1/√2) + (1/√2)² = -½
(x - 1/√2)² = -½
x - 1/√2 = ±√(-½) = ± √(-1) / √2
x = 1/√2 ± i/√2
√i = 1/√2 (1 ± i)
since it is a power of i, it lands on the unit circle on the complex plane
@@alexcarpentier5698 That's really interesting, I never realized that before.
so -1, I, and sqrt(i) are all on the unit circle. I wonder if the fourth root of i is as well.
3:20 from these equations you could have just done:
a^2-b^2=0 => |a|-|b|=0 => |a|=|b|
Then, knowing that you can go to the next equation:
2ab=1 => ab=1/2 => sign(a)=sign(b)
(Either both positive or both negative, because their multiplication results in a positive number)
Then, because their signs are equal *AND* their absolute values are equal, you can assume their both equal.
So now you have 2 solutions that differ by sign:
a=b=+-sqrt(1/2)
Everybody is like 'there' s an easy way:' and then has full paragraphs of calculations.
Just think in polar coordinates and rotations and the answer is obvious
Right? It should just be "what rotation composed with itself brings you to where i sits (90 degrees)". Boom 45 degrees. Boom, express as cos(45d)+i sin(45d)
Yep, I got it that way in seconds. When I was a teenager I was into the Mandelbrot/Julia sets, and the complex plane became my main jam.
@@AlanCanon2222 r/iamverysmart
MrFierMath what I did, like sqr of i is equal to sqrt -1, leave that sqrt and make -1 to polar, and then the moivre method and that is right?
I didn't have good teachers so this is the first time I understood this out
i don't care about the haters and their circle formulas, using various methos is really useful and gives more options, i know that formula but using this for fun is really nice
There is a method for determining square roots which uses a system of 3 equations on module, real part and imaginary part. It wasnt fully used here, and wasn't relevant in this case. This guy sucks at math and saying so doesn't make anyone a hater, just a skeptical person who knows a tiny bit about math
Clearly this guy sucks at math even though he got the right answer
So you just proceeded to state that there is such a method and didn't explain the method or at least name it, then you said he sucks at math even though he got the right answer in simple steps. Nice.
5:06 I noticed that not only are +1 and -1 solutions to a^4 = 1, but i and -i are too. More generally for a^n = 1, all nth roots of unity are solutions. Not that ignoring these solutions matters in this case, as they essentially just switch a and i*b but give the same final answers.
I dont think we need to consider a or b being i or -i since theyre respectively the real and imaginary parts of the solution were looking for, therefore a and b are real numbers
Then shut up
a and b are real they can't be i or -i
@@sebm2334 exactly
That was interesting to follow, I've forgotten so much math including basic algebra, this was very helpful and you did a great job of explaining it all.
English is also not my motherlanguage and i understood you justy fine, no complains, ignore these morons :D, they should be thankful you didnt speak your language and just put subtitles
Gmod2012lo1 english is not my mother language either, so it got me thinking: what if non-native english speakers understand other accents better when compared to native speakers?
@@JoaoVictor-gy3bk We really do. That's why I prefer being a non native English speaker.
@@JoaoVictor-gy3bk that doesn't make sense, nor is it true
He was speaking bamboo English. Its when I get up in the morning and the memory is loading to the ram but it takes a short time so I'm incoherent in the mean time.
What is your mother tongue? I assume it is west Germanic because you made the typo "justy" which reminds me of German richtig. German? Dutch? Frisian? Afrikaans?
Another way to solve the fourth root of four being equal to four is this:
Sqrt(Sqrt(4))=x
Sqrt(4)=2, so
Sqrt(2)=x
He didn't even need to do all that things for finding A, it just required the simple observation that a = b or -a = -b
This would mean 2a² = 1 and a = ± 1/√2
The Best way for me is to think of complex numbers as vectors on a complex plane. And if you raise an imaginary nomber to some power you make a rotation.
i^2 = i^(1+1) = -1 hense you make a 90 degree rotation anti-clockwise and went from the imaginary axis to the real one.
i^0.5 = i^(1-0.5) means that you make a 45 degree rotation clockwise. Now you only need to decompose your vector.
The real part is cos(45 deg), and the imaginary one is i*sin(45 deg).
@@lc1777 Yes, substituting a=±b from the first equation saves about five minutes of algebra compared to substituting b=1/2a from the second.
@@lc1777 I had the same approach. Much easier.
@@beeble2003 yup
Actually it can be done way simpler, if we know on the complex plane multiplication means that the absolute value of complex numbers multiply and the angle to the Real axis add up,
So i is 90° with an absolute value of 1
sqrt(i) is 45 ° with absolute value 1, use some cos and sin to get the actual values if necessary
it can also be 225° since you're going for 2*α mod 360° = 90°
MrRoyalChicken Actually, there is whole spectrum of solutions. They are
e^(i*pi/4 + k*pi), where k is a whole number (positive or negative, or zero)
niklas schüller So there are infinitely many solutions ;)
Luka Popovic out of all of those infinitely many solutions only two lay between 0° and 360° all other solutions are only a different way to reach those two points.
coming from the other standard form of a complex number: z = a + i * b = r*e^(i * phi + 2*pi*k), with r = sqrt(a^2 + b^2), phi = atan2(b, a). For the complex number "z = i", r=1, phi=(pi/2), so i=e^(i*(pi/2+2*pi*k)), i^1/2 = the result above.
6:31 isnt an easier way of doing that rearranging equation a² - b² = 0 to a² = b² in which case a = b
then substitute a or b in the second equation so 2a² = 1
rearrange so a = +-√2/2
same doubt
It can be a=-b
If a² = b²
a≠b
Never make that assumption
But it can be a = -b
Or
b = -a
@@ampleman602 a = -b or b = -a is the same. And he's right, it can only be a = b because of the second condition 2ab = 1, that only works if a = b. If a = -b you would get a negative output.
I’m not even doing this for school. I’m just interested.
same
Lucky
True
Same
Same, I'm really into this while i cant comprehend every step but i can feel his teaching energy pulling me into it.
I like the more unit circle method, where squaring imaginary numbers doubles the angle to the positive real line, so square root must half. i is 90°, thus it's square root must be 45°, the coordinates of which are ((√2)/2,(√2)/2), or (1/√2)+(1/√2)i
my nigga wearing a supreme shirt
real shit
هذا المثال في المنهج الدراسي في دولة العراق وهو من أبسط الامثلة في المرحلة الاخيرة (السادس الاعدادي) ❤
I like how you hold your microphone the whole time.
thanks!
And two pens in the other hand. That blew my mind.
honestly though, it just truly added to the enjoyment of the video. He shows so much excitement when explaining and it altogether created a really good video
i may be complex, but 1 is still the loneliest number.
Just 2 punny!!!
This shit got me sad asf.
But can't debate bt being lonely he is also the glory coprime
I isnt complex
at least he's positive about it, poor -1, lonely and negative :(
bro has infinite blackboards
🎅🎅🐡🥕
does anyone else notice how oddly motion blurred he get when he's on the edge of the screen, and only the edge
the camera's not in focus?
yesss I've noticed
@@kzushii it does not look like because it just seems out of temporal sinc, not spatial distortion
6:50 from there on it’s unnecessary, because you already got the equation a^2=b^2
Did anyone realise he is holding an Ood translator sphere?
Dem Rottensoul u realized too late. ruclips.net/video/STQPuHCiR8Y/видео.html
it's obvious when you write i as e^i*pi/2 then you consider the sqrt as power 1/2 then the sqrt is e^i*pi/4
The answer is squirt
The other solution doesn't fall out that way, but on the other hand there are infinitely many solutions in polar form.
There is also a restriction put into place when you remove the denominator of 4a^2,
4a^2 ≠ 0
=> a ≠ 0
Since none of the answers given were 0, it didn’t matter in this case, but you can’t simply forget the restriction.
This can also be done using re^iθ polar coordinates, where i = 1*e^[i(pi/2)], taking the square root yields sqrt(1)*e^[i(pi/2)/2], which is equal to 1*e^[i(pi/4)], in this case would be a vector 45° from the horizontal, with a length of 1, with the [ Re, Im ] coordinates being [ 1/sqrt(2), 1/sqrt(2) ], or in an equation, sqrt( i ) = 1/sqrt(2) + i / sqrt(2)
In polar form, the answer is very obvious. Half of a 90° rotation to the left is a 45° rotation to left or a 135° rotation to the right, which is the exact same result you got.
Paul Paulson oh that makes sense. If the square root of negative one is represented as 180 degrees, then the square root is equal to a one half power, meaning the power controls the rotation (in a way). 1 is 0 degrees so yeah.
The square root of negative one or i would then be 90 degrees
Cody Fan Yes, because with complex numbers, what is called "multiplication" is actually just the application of a rotation and a scaling. To square a complex number is just to apply the rotation and scaling twice. i is a rotation by 90°, while 2 i is the same rotation but also scales by a factor of 2. If you have e.g. 3 + 2i, you would have to convert it first into rotation and scaling. But just imagine it as the point (3,2) in the complex plane, draw a line to (0,0) and the length of that line is the scaling factor and its angle to the horizontal axis is the rotation.
Actually, to me, this is what complex numbers are really about and what their true meaning is. The thing with the imaginary square root of negative one is just some kind of mathematical trick (which is still usefull because it's often easier to calculate with it).
Paul Paulson It would be amazing if you could do a video where you illustrate this magnificent explanation.
Ayoub Merzak ruclips.net/video/mvmuCPvRoWQ/видео.html
You should watch this.
Just solved this today in my classes, takes less than a min if u do it by converting it into eulers form
i = e^(i*pi/2), sqrt(i) = e^(i*pi/4) = (1 + i) / sqrt(2). To get the other root, remember that i = e^(i*5*pi/4) and follow the same process
ikr!
I think a more interesting approach is by putting I in the form of an immaginary exponential with the argument written with the +2i(pi)n
Uh, easier to do the other substution no ?
a²-b² = 0 gives a = +- b
Plugged into 2ab =1 we get
2a² = +- 1 and so a² = 1/2 since a is real which also means a = b
So a = root(1/2) and b = root(1/2)
Or both negative.
This is better. In Math Simpler is Better.
More direct = simpler = more elegant
No danger, it's simple algebra. This is the more elegant way to finish this line of reasoning.
I agree but is a matter of taste. You need to add the idea that they have the same sign based on the fact that 2ab is one.
Stil the long route. It's easier to write i=e^(0,5*pi*i+k*2pi*i). Then the square root just halves the angle. So Sqrt(i)=e^(0,25*pi*i+k*pi*i). Basic angles from sine and cosine can translate back to the answer he got.
u can also write in polar form
sqrt (i)= i ^(1/2)= (e^(pi/2*i))^1/2= e^(i*pi/4)
now, write is as cos(pi/4) + i*sin(pi/4) and u get the answer ...✌
The area of the rectangle connecting i, -i, sqrt(i), and -sqrt(i) is sqrt(2)
Beautiful.
in a^2-b^2=1 you can simply add b^2 to both sides to get a^2=b^2 and then cut the roots then use this to get a=b then in 2ab=1 divide both sides by 2 to get ab=1/2 and then use a=b to get a^2=1/2 and then root both sides to get the answer to a and therefore b
i love how happy he is when talking about it
Very understandable, couldn't be more detailed.
And nice that you already did a video on the polar way ;).
Keep up the good work.
I admire this guy's ability so much. I lack mathematical knowledge, but it's so inspiring to see him perform this well with numbers. God bless you sir
It could be done in seconds using euler and polar form of a complex number
√i=e^iπ/4=cos(π/4)+isin(π/4)
=1/√2+i/√2
When you 4th root the 4, don't you get 4 roots which include 2 complex roots?
The other 2 roots are - +sqrt(2)i so the solutions just become mirrored.
a and b were defined as Real numbers
sweater is fresh my g
Thanks!!!
my only question is: How does 1/ {2(1/√2)} simplify to 1/√2 ?
Agreed. That simplifies to [sqrt(2)]/2.
@@crhodgkin 2/sqrt(2) = sqrt(2) ==> 1/(2/sqrt(2)) = 1/sqrt(2).
to show that sqrt(2)/2 = 1/sqrt(2); multiply top and bottom of LHS by sqrt(2) to get 2/(2*sqrt(2)), ,cancel the 2s to get 1/sqrt(2)
@6:13-6:14 you can hear an excited giggle as he encapsulates quartic root of 1/4th lol
That is the most beautiful part of this demonstration
You have very interesting questions you solve. The bits where you rationalize to create a very simplified fraction is genius. Did you practice stuff like this a lot to have such a deep intuition? You make it look so easy and natural. Thanks for these vids!
Agreed!
De Moivre's therorem: Am I a joke to you?
Therorerorerorerorero
What is it??
your energy is so uplifting. We love you. Keep doing what you're doing, with that little smile of yours you'll get anywhere :)
He has no energy. SJW's love his gimmick of holding a weird mic.
There's another way which I guess maybe is less lengthy
i= (2i/2)
=( i²+1+2i)/2
=(i+1)²/2
Hence sqrt i equals (i+1)/sqrt 2
No need to use a and b
Smart guy: reads the title as "sqrt(1)
Me: reads the title as SKRRRR
squirt)
SQUIRT
Actually is "sqrt(-1)"
Tzar C. Incorrect. That would be sqrt(i^2) - not sqrt(i).
In my dirty mind is squirrrrtttttttt
Use De Movire's theorem. First transform √(i) to polar form: cos(2nπ+(π/2))+isin(2nπ+(π/2)).
√(i)=(cos((4n+1)π/2)+isin((4n+1)π/2))^(1/2)
=cos((4n+1)π/4)+isin((4n+1)π/4). n=0, 1
=(1/√(2))+(1/√(2))i, or -(1/√(2))-(1/√(2))i
In the complex plane, when multiplying numbers, you add the angle and multiply the magnitude. Magnitude of i is 1, and i is at 90 degrees to real plane, so the magnitude of the sqrt is 1 and the angle is 45. If you draw on the complex plane a line at 45 degrees and give it magnitude 1, it becomes obvious that the end point of the line is 1/sqrt2 + i/sqrt2. And the same goes for a line directly opposite it
Yea, I thought the same. Why's noone talking bout that solution? But anyways, the video was interesting.
That is based upon DeMoivre's Theorem.
Isn't it much easier to have a look at the situation in polar coordinates, using the complex e-function? That way you see right away that there are two solution, each one a vector in the complex plane with length 1, and theta either being pi over 4 or five pi over 4, which then can be rewritten in regular form using Euler's formula.
i understood everything but i would never come with an idea like this
@@super_super_super485 😂😂
@@super_super_super485also 2/2...😂😂🤣and infinite can be made xd
Fax
basically :
45 degrees + 45 degrees = 90 degrees
or -135 degree - 135 degree = -270 degree = 90 degree
use radians u peasant
@@BlowitchAlt lolol
a²+b²=0 --> a=b
2a.a=1 --> 2a²=1 --> a²=1/2 --> a=+-1/sqrt2 seems another easier way to do that function system
My thought exactly. Makes me wonder rif he was just trying to prolong it for some reason. I like this channel, but this felt like it was intentionally taking the long way around when on the 2nd board.
You can use the complex exponential function sqrt(exp(i(pi/2))=sqrt(i)=exp(i(pi/4))=cos(pi/4)+isin(pi/4)=1/sqrt(2) + i/sqrt(2) 😅
When I studied maths we learned De Movire's theorem which allowed us to solve this much faster. But it's always fun to see a nice algebraic solution.
Just solve It in the algerbic way? Nah? trigonometric representation?? Much easier..
cos(π/2) + i sin(π/2) = i
√i has 2 roots:
Z1 = cos(π/4) + i sin(π/4)
Z2 = cos(-(3 π)/4) + i sin(-(3 π)/4)
yeah, I agree it's much easier
I wrote sqrt(i)=sqrt(e^(i*pi/2))=(e^(i*pi))^(1/4)=(-1)^(1/4)
It is a lot easier, but it also requires a lot more background knowledge.
+- (1+i )*(1/2)^(1/2) +2k pi
Well what about cos(9pi/4) + i sin(9pi/4)
"sqrt" being sus
Very thorough, I enjoy your problem-solving videos, and any Algebra student should be able to follow your steps. However, most Algebra texts would want you to rationalize the denominator (I personally am just fine with radicals in the denominator) and with so many computer video courses and testing, there's a good chance a computer would mark that wrong! Teachers today many times don't see the students' work which frustrates my daughter since of one dyslexia error producing an incorrect computer entry you can get zero points credit...
Why not take i=e^(i*pi/2) and then proceed...
My guess is, very few people that understand elementary math, don't know Euler's equation, or don't understand it well.
Though I'm missing a few things in this tutorial, for instance, he mentions the +/- later on, but with the first squaring, squaring (sqrt(i)), he skips over it (he doesn't explain why he doesn't look at -sqrt(i). But I like the very basic and simple deduction any high schooler an follow ;). Still, it would've been nice to get some more insight on the surprising result, like that it's related to cos(45).
@@Misteribel
Actually, the two answers he gets are only differing in a minus sign. So you could say that sqrt(i) = +- (a+bi), where a = b = 1/sqrt(2). So -sqrt(i) isn't really _skipped._
Btw, regarding the connection to cos(45) (and also sin(45), since they both are 1/sqrt(2)): Whenever you take the square root of a complex number on the unit circle the argument (angle) halves. Since i is represented by an angle of 90 degrees (pi/2) on the unit circle, i.e. i=e^(i*pi/2), the square root of i is represented by 45 degrees (pi/4) on the unit circle, i.e sqrt(i) = +- e^(i*pi/4) = +- [cos(pi/4) + i*sin(pi/4)] = +- [1/sqrt(2) + i/sqrt(2)] - which is exactly the result he gets in the video.
I realised I was rambling a bit, and I just did it the way Sourav Ganguly described... Oh well. :)
@@Sjobban112 exactly. You explained my point in better detail, thanks. And I should have said 'skimmed over', since this is introductory, leaving it out may leave people wondering. I didn't mean to say he was wrong here.
@@Misteribel
No worries. Cheers! :)
身為高下課且很會教的高中數學老師，回家原本是想想聽聽外國萌子的asmr+whisper，讓我放鬆休息一下，順便做些別的實情，哪知話鋒一轉。連續播放就幫放那你這了！原本想說那就結束，結果下一個又推薦Eddie Woo ～～～～喔喔 我希望 我未來也可以用英文教數學！！！！～
reading the comments section made me feel dumber every comment
I believe that your speaking way is not important as long as you explain very well ; thank's for helping me again
4th root of -1?
edit: got this by doing (-1^1/2)^1/2 which equals -1^(1/2*1/2) which equals -1^1\4 or the 4th root of -1
-1 can be written as i^2. Substituting we get ((i^2)^1/2)^1/2 which is the same as i^1/2 or the square root of i.
You can actually think about it easier with polar form.
z=sqrt(i) is the same thing as z²=i.
Representing z in polar form is re^it with t being the argument of z, or theta. The square of this would be r*r*e^it*e^it. r*r is r² and e^it*e^it is e^2it. The magnitude of i is 1, so r² is 1, and since the magnitude of a complex number is always positive, r=1, so the square root is on the unit circle. The argument of i is pi/2. However, you can't consider only one solution, since the argument resets to 0 at the real number line. The equation isn't just 2t = pi/2. It's 2t = pi/2 (mod pi). That means 2t is between 0 and pi.
There are only two solutions to that, t = pi/4 and t = 5pi/4. So the two solutions are e^ipi/4 and e^i5pi/4. The two solutions lie on the y=x line, so 2x²=1. x=±sqrt(2), and so z = ±sqrt(2)±isqrt(2).
7:00 Alternatively, you could use a^2 - b^2 = 0 to get a^2 = b^2, thus b has the same two possible values as a.
Edit: nvm
You can't because b can be 1 and a -1 and there a^2-b^2 will still equal 0 even if a is the opposite of b, a^2 - b^2 = 0 means that a and b have equal magnitude but equal or opposite values
@@mhmd-mc113 Oops, you're right
I voted this video up. But I wished that you also showed a geometric solution. When you are used to seeing complex numbers on a plane and you know how they multiply, you can solve this in your head pretty quickly. However, doing this in my head, I forgot that there are two answers.
Jim Greene
I did. You can find the link in the description.
7:33 how is 1/2(1/sqrt(2))=1/sqrt(2)?
Can someone explain please?
Hello! you could think of 2 in the sense of sqrt(4) when multiplying to 1/sqrt(2). sqrt(4)*1=sqrt(4) so that whole thing is sqrt(4)/sqrt(2). Just like how sqrt(x)*sqrt(y)=sqrt(xy), its the same with division, sqrt(x)/sqrt(y)=sqrt(x/y), so sqrt(4)/sqrt(2)=sqrt(4/2)=sqrt(2)
Hope this helps!
@@Squ_red Ohhh I get it now! Thanks for the explanation!
Euler's formula just makes everything easier
My first instinct was de Moivre’s Theorem, which of course works out very nicely, but I appreciate this unique algebraic approach. I love seeing how different people come up with completely different ways of solving math problems!
I recommend this maths problem
ruclips.net/video/z2OyVIJznHw/видео.html
This is solving a simple problem in the most roundabout way possible.
On a complex plane i is 1in length on a 90° angle
Solution to ✓i is the number that ² to I
So 1/√2 in length, with 45°& 225° being the only possiblities
45° & 225° amount to 1/✓2(1+i) & 1/√2 (-1-i)
For complex number operations in multiplication, division, power, and root, covert them into polar form then use DeMoiver's theorem. i=cis(pi/2) => sqrt(i)= cis(pi/4) and cis(3pi/4)=1/sqrt(2)+1/sqrt(2)i and -1/sqrt(2)-1/sqrt(2)i.
If one understand geometry of complex numbers, it is immediately obvious that solution is just a pair of unit vectors, one with angle of 45° and one with angle of 45° + 180° = 235°.
Yaaaa I was gonna write this comment haha, can solve using demoivers in about 10 second
We can do it in our heads by applying Pythag' on the complex plan, mate.
(But I admit that I missed the second solution - I forgot that i has a polar coordinate angle measured _clockwise_ from the positive reals.)
I also did it visually and missed the second solution.
No, the second solution is not a conjugate (measuring clockwise from the reals), it is a negation (measuring from the negative reals counter clockwise)