be careful when an imaginary number is raised to a fractional power

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3 жыл бұрын

Do you like imaginary numbers and complex analysis? Take a course from Brilliant via brilliant.org/blackpenredpen/ and start learning something new. That link also gives you a 20% off discount on their premium subscription.
The powers of the imaginary unit i are always intriguing. Here we will examine i^(4/4). Is the result 1 or i? We have to be careful when an imaginary number is raised to a fractional power!
0:00
0:06 is i = 1?
4:23 (i^4)^(1/4) vs (i^(1/4))^4
10:14 (i^3)^(1/4) vs (i^(1/4))^3
13:53 summary on z^(m/n)
15:41 check out Brilliant to learn more!
16:35 bonus part
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Пікірлер: 702

@official-obama 2 жыл бұрын

NEVER calculate roots without a trusted adult's supervision

@blackpenredpen 3 жыл бұрын

Good morning! Bprp, 4:55 am!

@neilgerace355 3 жыл бұрын

Good morning from 15 hours in the future!

@kakashiuchiha1077 3 жыл бұрын

-2 = (-2)^2×1/2 = √{(-2)^2} = |-2| = 2 Which step is wrong plz tell me???

@jonatanpinsard5811 3 жыл бұрын

Good moorning! Here is 9:10 am

@8DJYash 3 жыл бұрын

It's evening here

@epsilia3611 3 жыл бұрын

@@kakashiuchiha1077 -1=i² as far as I know, so you might use that property in order to get a result. Also, the property "a=sqrt(a²)" is true only in certain conditions on the a. It's just like when you use a theorem, you want to have the right conditions in order to use the theorem (for example, you use the Pythagoream theorem only when you have an angle of pi/2 at one of the angles of your triangle)

@NatoSkato 3 жыл бұрын

Its like saying 1 = sqrt(1) and then picking -1 for the square root and then saying 1 = -1

@user-tb8lg5jo1u 3 жыл бұрын

but sqrt(1) is not equal to -1

@NatoSkato 3 жыл бұрын

@@user-tb8lg5jo1u then what is -1 squared

@user-tb8lg5jo1u 3 жыл бұрын

@@NatoSkato its not how it works my friend

@Felixr2 3 жыл бұрын

@@user-tb8lg5jo1u You understand what he means. And if you don't, here's the exact same thing worded slightly differently. (-1)^2 = 1 sqrt(1) = 1 therefore, -1 = 1

@9merk_ 3 жыл бұрын

I think this might be a bit more clear: x = sqrt(1) = 1 x^2 = 1 |x| = sqrt(1) x = +- 1 Then concluding 1 = -1

@yoavboaz1078 3 жыл бұрын

you can do a real number version of it -1=(-1)^1 -1=(-1)^(2/2) -1=((-1)^2)^(1/2) -1=1^(1/2) -1=sqrt(1) -1=1

@yoavboaz1078 2 жыл бұрын

@@gtgagaggagagagga the step above implies that

@DeJay7 2 жыл бұрын

Yep. Whereas if you raise it to 1/2 firs- oh Edit: It does work though. You get -1 = sqrt(-1)^2 which is not real but by definition it is -1 = -1 (real)

@rythmx123 Жыл бұрын

you always have a ± in front of a square root which implies you have 2 solutions (-1,1), since you have -1 on the left hand side you're saying a negative value is equal to the positive with is not possible, thus, we discard the positive value and take the negative one, for everything, negative = negative, so -1 = ±√1 = -1

@jtris01 Жыл бұрын

@@rythmx123 The reason we add a ± in front of square roots is not for the reason you think. When taking the square root of x², you must consider that x² = |x|². Suppose we have the following equation: x² = 16 |x| = 4, or x = ± 4 The reason we get the issues in the original problem is that (-1)^(2/2)= (-1^½)² = i².

@yf-n7710 Жыл бұрын

@@rythmx123 Yes, exactly. There's a hidden solution. It's exactly the same as the problem in the the video, where there are three hidden solutions. It's just that for some reason +/- in front of square-roots is more automatic than (+/-/i/-i) in front of fourth-roots

@vibby1004 3 жыл бұрын

Step 3 creates 3 extraneous solutions, it’s really cool imo

@guitarninja0403 3 жыл бұрын

Åland Islands 🇦🇽

@ruffifuffler8711 3 жыл бұрын

Can only spread gravity, but never factor it, it can be morally partitioned by intervention, and reduced, .. .but never factored. . After the euler plunge, ...and before coagulation & later spin, ...pre-podal "i" assumes the identity of the singularity operator. By giving it exponents, one removes it from singularity parity, ...and it then it fractures because of the time necessary to give it new meaning.

@nanni5230 3 жыл бұрын

@@ruffifuffler8711 What

@ruffifuffler8711 3 жыл бұрын

What? "i" is the 1st instance of gravity.

@cougar2013 3 жыл бұрын

I believe the real reason is that the root function is not analytic in the complex plane

@Sam_on_YouTube 3 жыл бұрын

That use of quotes is called "scare quotes." It is a play on words from how they are used when speaking, which is as "air quotes." When you use quotes for something that isn't literally being said it indicates that you are using the term, but not attributing the meaning to yourself. So you are saying it is okay, but you are distancing yourself from the word. You are putting it in quotes so it is not attributed to you, but just to what other people might say.

@QweRinatrtY 3 жыл бұрын

"you" are "putting" it in "quotes" so it "is" not attributed "to" you, but "just" to what "other" "people" might "say"

@kiro9291 3 жыл бұрын

it's for quoting other people's usage of the word, but emphasizes that it's how others use it

@RubyPiec 3 жыл бұрын

"So" "if" "someone" "copyrights" "every" "word" "in" "the" "dictionary" "this" "is" "how" "to" "avoid" "getting" "sued"

@Sam_on_YouTube 3 жыл бұрын

@@RubyPiec It was tried recently with music. Someone generated every possible combination if notes and stored it on a hard drive. Every song below a certain length is on there. Tried ti copyright it. Got rejected.

@DarthNihilusKorriban 2 жыл бұрын

ah, yes, "Reapers"

@tylershepard4269 Жыл бұрын

As an electrical engineering PhD student, this phenomena is what we refer to as “phase unwrapping.” Where the phase is the argument of the complex number. The wrapped phase always lies between -pi and pi. Once the wrapped phase goes above pi, it “wraps” back around to -pi and continues increasing. To unwrap the phase, look for discontinuities in the phase and add 2*pi to the phase thereafter for each discontinuity encountered.

@heinrich.hitzinger 8 ай бұрын

Something similar happens when you integrate a complex function with a discontinuity...

@lunstee 3 жыл бұрын

I find this reminding me of derivatives and indefinite integrals. The integral has an arbitrary constant that the derivative is blind to, so the derivative of the integral of a function is the original function. The integral of the derivative of a function on the other hand, can differ from the original function by a constant. The denominator of the exponent adds a similar ambiguity, which the numerator of the exponent removes if given the chance. Applying the numerator first wastes its many->1 mapping, giving you many values if you do that before considering the denominator. In both cases, we're applying some operation (whether differentiating, or raising to the power of n) and the "inverse operation" (integrating or raising to power of 1/n respectively) , in one order or the other. The gotcha is where one of the operations involved is a many->one mapping: the inverse mapping is necessarily one->many. Things are tidy when you let the operation clean up after its own inverse, and messy when you don't. The most primitive many->one operation is of course multiplying by zero. The inverse mapping would be dividing by zero. We're well conditioned to not divide by zero, but can have all sorts of fun trying to conceal where it happens. It's almost like hiding vegetables in kids' food. All sorts of funny stuff then happens when you end up multiplying things by 0/0.

@zeno4253 3 жыл бұрын

Yep I think you nailed it

@watwat7050 3 жыл бұрын

It's always such a joy to see you solving these problems and explaining them clearly. Cheers mate!

@blackpenredpen 3 жыл бұрын

Thanks!

@stapler942 3 жыл бұрын

If I'm understanding correctly, this might be partly a consequence of the fact that z^x is not truly invertible, and the reciprocal of an exponent is not a true inverse of that exponent. Even in the real numbers, even powers are not invertible. Or perhaps we can say that fractional powers are not truly associative for complex numbers (or whatever the equivalent term for associativity is for powers).

@bartekabuz855 3 жыл бұрын

"who did that? Yes, it was me (DIO!)"

@joonjoon2368 2 жыл бұрын

I’m just mesmerized by his marker quick switch

@swisssr Жыл бұрын

In your second example, the order didn't matter for the ^3/4 because you considered all the possible values for i^3 by writing it in the polar form. But for the first example ^4/4, you just reduced i^4 to 1, instead of writing it in polar form (e^(i*pi*(1/2 + 2*n)))^4 , which gives i (as expected) when raised to 1/4. When you do the fraction i^(1/4) first, it works just because you use the polar form. In summary, as long as you write it in the polar form (not only when the power is fraction, but integer too), it works. The polar form also explains the "real number version" as someone mentioned in another comment: -1=(-1)^1 -1=(-1)^(2/2) -1=((-1)^2)^(1/2) -> wrong, you should write (-1)^2 in the polar form e^(2*i*pi*(1 + 2*n)), which when under sqrt, it gives -1 as expected. -1=1^(1/2) -1=sqrt(1) -1=1

@heremoanalau1351 Жыл бұрын

Wowww tysm 🙇

@damianbla4469 3 жыл бұрын

17:15 Why we should take firstly i^(1/4) and then ^4? Because "i^(1/4)" gives us four different results - but every of them, raised to the 4th power ("^4"), gives us the same one complex number - which of course is correct and is answer to our original question. Why we should NOT take firstly i^4 and then ^(1/4)? Because "i^4" gives us one result - but this reesult, after taking 4th root of it ("^(1/4)"), gives us FOUR DIFFERENT complex numbers - only one of them is correct and is answer to our original question. Writing it shortly : i^(1/4) --> four different results --> take them "^4" ---> one result, which is correct and is our answer i^4 --> one result --> take it "^(1/4)" ---> FOUR DIFFERENT :( results, only one of them is correct and is our answer

@camrouxbg 3 жыл бұрын

This is so clear! Thanks so much! Really refreshed roots of unity for me, and also refreshed some of those basic exponent "rules" that I hadn't fully internalized.

@Viewer2812 3 жыл бұрын

Yoo it's finally here.

@aashsyed1277 3 жыл бұрын

SAME FOR MEE

@nicolasgoubin 3 жыл бұрын

And here we are

@orionsrash1515 3 жыл бұрын

Quote of the year: "Just reduce & be happy!". Great video, thanks for the enthusiasm.

@HopeP0H 2 жыл бұрын

Took me a minute before I realized that there were multiple roots. This reminds me of some fourier transform examples I was thinking about, if I didn’t solve those incorrectly. What I mean by that is when multiple functions result in the same function after a transform, so, for example, when you apply a fourier transform to a function, and then do the inverse fourier, you could end up with a completely different function.

@SquibbyJ 2 жыл бұрын

I always enjoy your videos, even when I learn nothing new I love the way you explain math. You do a great job of going through concepts thoroughly and that’s very admirable

@blackpenredpen 2 жыл бұрын

Thank you!

@Rofida-rh1wq Жыл бұрын

Heeeeeey I literally don't know anything from the video except the (i)🥲 but i enjoy watching it even with that

@MrCigarro50 3 жыл бұрын

Thanks for this video. I am still taking my notes to understand it.

@RoderickEtheria 2 жыл бұрын

Mistake is ruling out the complex roots when square rooting.

@guillaume5313 3 жыл бұрын

Great video! Things I didn't know but aren't too hard to understand

@dr.rahulgupta7573 3 жыл бұрын

Excellent presentation. Thanks .

@kiro9291 3 жыл бұрын

that was a satisfying explanation, thank you

@joaopedroalves1200 3 жыл бұрын

That's literally something I was working with before. It makes difficult because of the fact that exponentiation is actually a multivalued function, but we usually define only one value for it. Given the nature of the pattern the solutions create in a sequence, I would really suggest the usage of a kind of "modulo operation" extended to complex numbers, so these problems could be avoided.

@createyourownfuture5410 2 жыл бұрын

Not very related but I had a doubt with absolute value. Is |i|=1? And is |1+i|=√2? Because on Wikipedia I saw that absolute value is the distance from 0. On Wolfram Alpha it says that the above statements are true.

@ThomasTheThermonuclearBomb Жыл бұрын

@@createyourownfuture5410 Yes, abs(i) = 1 and abs(1+i) = sqrt(2)

@createyourownfuture5410 Жыл бұрын

@@ThomasTheThermonuclearBomb thanks mate

@wanpengli-mr8vv Жыл бұрын

YES, the key is multivalued function

@xXJ4FARGAMERXx Жыл бұрын

@@wanpengli-mr8vv multivalued function that's not a function. That's a mapping.

@angelmontoyagallardo5679 2 жыл бұрын

This guy is escapism from reality and my reality problems are mathematical problems at school lmao and he actually does it rly well

@242math 3 жыл бұрын

very intriguing and expertly done

@SpaceWithSam 3 жыл бұрын

So interesting, great 👍 content!

@romaobraz4295 Жыл бұрын

this is brilliant. great vid!

@UniqueNCS 3 жыл бұрын

isn't it just because the formula (I ^n)^m = i^(m*n) is only valid when m & n are natural numbers?

@ryang628 3 жыл бұрын

If m and n are integers, and z is nonzero, then (z^m)^n always equals z^(mn). Otherwise, this law does not hold, except under particular combinations of restrictions on m, n, z. Coincidentally, I had a discussion about this just hours before this video was posted; search for "Regarding the necessity of the coprime condition" on math.stackexchange for my justification of what bprp said.

@cougar2013 3 жыл бұрын

I think it boils down to the fact that the root function is not analytic in the complex plane

@pedrosso0 2 жыл бұрын

It is valid for all n and m I think... However that's only for one solution, there are multiple solutions which means that although you could bring it down to i^(m*n) there are other sols.

@grrgrrgrr0202 Жыл бұрын

@@pedrosso0 It is simply NOT valid unless m and n are integers, or unless l is a positive real number (and for the latter, you have to make sure that the logarithm you use extends the usual real one).

@XJWill1 10 ай бұрын

(z^w)^p ? z^(w*p) The question is: when are these equal. The answer is: (1) If p is an integer OR (2) If -pi < Im(w * Log(z)) 0

@jithin.kesanapalli8317 3 жыл бұрын

Wow,great sir you are helping us with common contradictions in mathematics

@MathElite 3 жыл бұрын

Morning 8:30am EST Love your videos

@akshatjangra4167 3 жыл бұрын

Hello there!

@robfrohwein2986 7 ай бұрын

Thanks again... Good explanation 😀

@miguelalejandromorenobarri4759 Жыл бұрын

You, sir, are one of those who have best explained this topic.

@SuperYoonHo 2 жыл бұрын

Awesome! lots of thanks

@angelmendez-rivera351 3 жыл бұрын

Once again, this gets into the whole issue where everyone confuses "roots of a polynomial" with "the nth root of a number" and think that they are the same thing. They are NOT. The roots of a polynomial are a multiset. The nth root of a number is given by a *function.* Symbolic expressions are always *necessarily single-valued:* that is just the way mathematical notation works. We do not perform arithmetic with multisets, the concept is just short of nonsensical. The same applies with complex exponentiation. It IS true that z^4 - 1 has 4 roots, with the multiset of roots being {1, i, -1, -i}. HOWEVER, when we talk about the 4th root of 1, this has *only 1 answer.* Why? Because the 4th root of a number actually has *nothing to do* with the roots of a polynomial. It is a *FUNCTION.* That is what the radical symbols and these fractional exponents denote. Having noted this, it is important to note, that for complex numbers x, y, z, the so-called "identity" (x^y)^z = x^(y·z) is just a widely-believed myth, and is false. It certainly is true if y and z are integers, but otherwise, this is rarely true. So rather than relying on a myth, you need to rely on the *definition* of exponentiation. For nonzero x, mathematicians usually take x^y to mean exp[y·log(x)], where exp has a precise, unambiguous definition, and log(x) = ln(|x|) + atan2[Im(x), Re(x)]·i. And using this definition, it becomes clear, that in general, z^(p/q) = [z^(1/q)]^p and NOT z^(p/q) = (z^p)^(1/q). And as this video has clarified, (z^p)^(1/q) = z^(p/q) only when |gcd(p, q)| = 1.

@angelmendez-rivera351 3 жыл бұрын

Also, while it is insidious to write an expression like i as i^(4/4), it is not actually wrong. It is perfectly valid mathematical notation.

@angelmendez-rivera351 3 жыл бұрын

@@christiansanchez7614 A multiset is like a set, but repetition of elements is meaningful in a multiset, so multiplicity does matter. For example, the set {2, 3} and the set {2, 2, 3, 3, 3} are the same set, because the repetition of the elements is actually irrelevant. But the multiset }2, 3{ is different from the multiset }2, 2, 3, 3, 3{, because in the former, each element has multiplicity 1, while in the latter, 2 has multiplicity 2, and 3 multiplicity 3.

@mathsman5219 3 жыл бұрын

Thank you so much.❤️

@XxfishpastexX 3 жыл бұрын

thank you 😊

@williamestey7294 3 жыл бұрын

Well explained! Curious what do you do?

@SomeTomfoolery Жыл бұрын

Man that 3i joke really got me chuckling, thank you

@paradoxreboot 2 жыл бұрын

i havent done complex analysis in a few years but the proof of that last theorem is going to keep me up tonight

@JG-ne2mx 3 жыл бұрын

“Yes it was me 😁” great line

@ManuelRiccobono 2 жыл бұрын

This video explain exactly why i hate math. It is a language where you have to follow strick path, because the moment you try to take a jump to the nearest path, you fall into the abyss of wrong answers

@BryndanMeyerholtTheRealDeal 5 ай бұрын

2'43 kind of like the whole i^n package

@remo9229 3 жыл бұрын

I always have a similar problem stuck in my head e^(2pi*i) = e^0 ln(e^(2pi*i)) = ln(e^0) 2pi*i = 0 2 = pi = i = 0

@manojsurya1005 3 жыл бұрын

The conclusion at end was great

@jeremysharpe5467 3 жыл бұрын

-1 = -1^(2/2) = (-1^2)^(1/2) = 1^(1/2) = 1 And similarly, we see that that (-1^2)^(1/2) has 2 answers, while (-1^(1/2))^2 has 1 answer only (-1).

@ieatgarbage8771 2 жыл бұрын

Is there any notation for + or - that includes rotations other than 180 degrees? Seems like it would be helpful whenever you’re dealing with a root above a square root

@kekt536 2 жыл бұрын

my favourite video so far. very fun

@LuisBorja1981 3 жыл бұрын

This kind of false paradoxes are a great way to explain the concept and importance of the property of injectivity and bijections.

@popcorn485 3 жыл бұрын

Cool. Running this by my A Level students tomorrow!

@hypergaming2051 3 жыл бұрын

Very good explanation

@arnavoza527 3 жыл бұрын

I was so excited when I thought i is actually 1😅😅

@pronounjow 3 жыл бұрын

I was like "Not this again". lol It's a similar situation to multiplying (-1)^(1/2), which is i AND -i, by itself.

@MathIguess 3 жыл бұрын

In terms of distance, you can see i as some kind of 1

@pronounjow 3 жыл бұрын

It's 1 at a different angle, or orientation. For example, 1 is really 1@2nπ radians, and i is 1@((π/2)+2nπ) radians, where 'n' is any integer you choose. You must take both dimensions into account, which is why BPRP used Euler's Formula to show the truth behind this oddity. Euler's Formula connects the number 'e' with cos and sin, with cos and sin each representing a dimension.

@justfrankjustdank2538 2 жыл бұрын

he would look so different and the same without the beard

@jibster5903 2 жыл бұрын

Know: The nth root has n solutions in the complex world, so quadroot(1) = 1, i, -1 or -i. This is because of a really nice geometrical property in the unit circle but i'll leave the visualisation up to someone else

@marcosolza3698 Жыл бұрын

Brilliant explanation! 😃

@TheJaguar1983 5 ай бұрын

I once decided to calculate the value of i^(1/2), and then i^(1/3). Upon realising they fit common trig identities, I looked up and was reminded of Euler's formula. That was a face-palm moment for me 😂

@alexandrentema6202 3 жыл бұрын

When "100 limits "come out in one video!! watching from Mozambique(Africa)

@jayska5802 3 жыл бұрын

Phenomenal video

@meatystalactite531 Ай бұрын

Intro almost gave me a conniption... thank you mister Pen :D

@iamnotuta2658 Жыл бұрын

I like how I interpreted the title as a warning that I might break reality computing this thing.

@sergeygaevoy6422 9 ай бұрын

It is something like a clock. 1) Iff gcd(m, n) = 1 so exists suck k so (m * k) mod n = 1. (x * m) mod n = y x = (y * k) mod n Otherwise you cannot revert the multiplication. (m * k) mod n is always divisible by gcd(m, n). 2) gcd(m, n) > 1 also leads to non-trivial zeros: m * (n/gcd(m,n)) mod n = lcm(m, n) mod n = 0 Obviously irreversible. 3) In this case we cannot revert the power.

@elaadt 3 жыл бұрын

Awesome stuff

@johnernst2472 3 жыл бұрын

How about delving into the intricacies of Mandelbrot fractals?

@cmilkau Жыл бұрын

This video needs more views. Even maths channels tend to ignore this fact.

@cmilkau Жыл бұрын

A common generalization of this error is confusing a^(1/n) b^(1/n) with (ab)^(1/n) while taking naive roots. For people who don't believe they're not the same, set a = b = -1 and n = 2. With naive roots you get i² on the left and 1 on the right.

@fghsgh 2 жыл бұрын

I feel like there is an insight to be gained at 4:38. i^4 is indeed =1, however... i^4=(e^i(π/2+2nπ))^4=e^(i(2π+8nπ)) this last angle, 2π+8nπ, is indeed 2π, which means i^4 is indeed 1, however, notice that the coefficient of n is 8, not 2 this means that e^i0π, e^i4π, and e^i6π are not included! so then, if you divide the angle by 4 again, you only get back to π/2, which is i, without the 3 other answers so if there is a way to keep track of how many times 2π you are going around the circle, this could be resolved

@_kopcsi_ 2 жыл бұрын

I guess the symmetry breaking (i.e. the violation of commutativity) of the power operator is due to the logarithm function and its issues with its branches.

@adityagunjal1951 3 жыл бұрын

Sir please make a video in which problems from book named "which way did bicycle go?" Are solved.

@asparkdeity8717 Жыл бұрын

The problem is the function z^k requires a branch cut for any non-integer k, specifically z^1/4 in this case; the point z = 0 is a branch point, which can simply be shown by the discontinuity of arg(z) when jumping from say 2*pi back to 0

@cirax856 11 ай бұрын

best explanation I can think of would be that if you have the roots, you can only choose the same ones for both sides, and by choosing i on one of them, you can't choose 1 on the other.

@quneptune 5 ай бұрын

also worth mentioning (-1)^2/2≠√((-1)^2) what you did is kind of an imaginary absolute value function

@davidwright8432 Жыл бұрын

Why didn't I think of that!? Sometimes, the 'obvious' takes a little longer, a little more effort! Thanks, BPRP!

@TitanOfClash Жыл бұрын

Please can you make a follow up video to this, because I don't understand where the necessity of reducing the fraction comes from. That's what I was waiting for the whole time!

@blackpenredpen Жыл бұрын

Oh, it's because if we dont reduce the exponent first, then we would run into the situation in the beginning of the video i=1

@TitanOfClash Жыл бұрын

@@blackpenredpen firstly, thank you so much for replying. I'm such a big fan! But secondly, I don't really understand *why* that's the case. I can see that it happens mathematically, but surely i^(4/4) should be exactly identical to i. And taking the fourth power first or the inverse fourth power should always be the same. Isn't it like a * b is always b *a ?? Many thanks

@rudrodeepchatterjee 10 ай бұрын

5:59 an easier way to prove that 1 ^(1/4) has four solutions is just to break 1/4 into 1/2 × 1/2. That way, we get (√1) ^1/2. √1 gives ±1. √±1 gives : √1=1, -1. √-1=i, -i So we get ±1, ±i as our solutions.

@petteripan2658 Жыл бұрын

Sir, you open my vision in maths world again 😲

@Setiny 3 жыл бұрын

The problem here is you should treat i^x as a single value function e^(iπx/2)

@nicolasgoubin 3 жыл бұрын

Looool i felt so small when he said "Here, let me tell you..."

@sergeygaevoy6422 9 ай бұрын

1^(1/4) is +1, -1, +i, -i (four values of course, one in each directions: east, west, north, south) so we can only assume that i is one of these four numbers.

@Alisayadi2000 Жыл бұрын

No one answered this question of mine in the university and they just confused the question, I was very upset at that time that no one would help me, everyone was just looking to get a grade from the specified material, learning the course was not important, this caused After some time, due to the unanswered questions and the inefficiency of the university, I became discouraged and eventually fired, while I loved learning the material from the bottom of my heart, I thank you for making the video and clearing one of my mental uncertainties. I wish others were like you in education❤️

@parkershaw8529 3 жыл бұрын

Between step 4 and 5. It's very easy to see within real numbers if you start with -2 instead if i.

@EvilDudeLOL Жыл бұрын

It is like saying 1 = 2 because (1)^2 - 3(1) + 2 = (2)^2 - 3(2) + 2. Just because 1 and 2 are the roots of x^2 - 3x + 2, doesn't mean they are equal. Similarly, x = 1^(1/4) [or x^4 - 1 = 0 if you prefer] works for both i and 1 as solutions but doesn't prove that i = 1.

@leftenanalim Жыл бұрын

When you perform the reverse operation to find, the answer is not just equal 1. It's equal to either 1 or sqrt of -1. This is similar to the sqrt of a number, where it has 2 possible answer. When you perform an even root, there's always more than one solution. And in this case, you have to choose sqrt of -1 as answer because you know the value of i initialy. I always point out the importance of not neglecting the multiple solution when one is performing an even sqrt. For example, sqrt of 4 is either 2 or -2. And you can see why this is important when you are dealing with problems like this

@ulftho Жыл бұрын

The mistake is to equate 1 with e^(i2n*pi). If we introduce polar representation and Eulers formula from the beginning as in i = e^i(pi/2 + 2m*pi) we will get i^4=e^i(2pi + 8m*pi) ruling out every n except n=1+4k, and thus preserving the relationship between 1 and i^4.

@sylv512 2 жыл бұрын

the beard makes your maths 10x better

@cougar2013 3 жыл бұрын

I believe the real reason is that the root function is not analytic in the complex plane

@AuroraNora3 3 жыл бұрын

Alternatively, you can avoid this ENTIRE MESS by restricting yourself to principal values, which is what any advanced calculator does. Wolfram Alpha, or any TI calculator, will show you that (i^4)^(1/4) = 1. The reason being: 1. The order of operations tells you that (i^4)^(1/4) = 1^(1/4) 2. The principal value of 1^(1/4) is 1... (NEVER i or -1 or -i) But how does your calculator choose the principal value? Before multiplying ANY exponents, it reduces the complex number z to its simplest polar form, with the constraint -pi < Arg(z)

@Cobalt_Spirit 2 жыл бұрын

So you could say that arg⁡(z)=Arg(z)+2kπ, k∈Z (arg(z) is just the argument, whereas Arg(z) is the principal argument)

@Cobalt_Spirit 2 жыл бұрын

But why would you only consider the principal fourth root of 1, instead of all of them? Is it because z^(1/4) is defined as the main fourth root of z and not the others?

@gtgagaggagagagga 2 жыл бұрын

@@Cobalt_Spirit arent x^n only principal roots?

@Cobalt_Spirit 2 жыл бұрын

@@gtgagaggagagagga Usually, yes. But not in complex numbers.

@gtgagaggagagagga 2 жыл бұрын

@@Cobalt_Spirit eh of course *facepalms

@user-gs6lp9ko1c Жыл бұрын

This problem also occurs taking rational powers of negative real numbers. Found a mistake in a graduate level optics text book (discussing effects of atmospheric turbulence where rational powers occur all the time): The author took a perfectly good integral from 0 to infinity and made the mistake of taking half of it from -infinity to infinity. Made the base in the integrand negative and blew up my numerical computational software.

@danielschwartz6795 8 ай бұрын

Ah yes, playing silly buggers with multivalued functions. Always have to be careful about that sort of stuff.

@myidanny 3 жыл бұрын

Why is n always 0, 1, 2, and 3? Because of the 4th root, so we need 4 solutions? So a square root would make us have n = 0, or 1?

@yelbuzz 3 жыл бұрын

Yeah and it's partially related to how e^(i * x) works as a rotation in the imaginary plane. Every time you increase x by exactly 2pi (or increase n by 1 if x is some value + 2 * n * pi), that value will remain the same since you're rotating 2 pi radians around the circle, which is the same as not rotating at all. That's expressed through the "+ 2*n*pi" that you'll often see when writing the numbers this way like he did at 6:36. Another way of saying this is that it has a period of 2pi if you were to think of it more like a wave, which it sometimes acts like. When you raise something like e^(i * 2*n*pi) to the power of (1/4), you get e^(i * (2 * n * pi) / 4) or e^(i * (n * pi) / 2). n represents any integer, and they will all give valid solutions, but now we can see that every time we increase n by 1, it increases x by (pi / 2) instead of the 2pi that it did before. In this case, we'll have different solutions for n = 0, 1, 2, and 3, and it's not until n = 4 that x would be increased by 2*pi, resulting in the same number as n = 0. If we were to take the square root, or an exponent of (1/2), increasing n by 1 would increase x by pi, so once you get to n = 2, it would be the same as n = 0. So yes, a square root would have n = 0 or n = 1.

@myidanny 3 жыл бұрын

@@yelbuzz this is awesome, thank you very much for taking the time to explain! This is very much appreciated

@oenrn 3 жыл бұрын

Because after that it loops around and you get repeating answers. n=0 gives you e^i0=1 n=4 gives you e^i2π=1 n=8 gives you e^i4π=1 Same for n=12, 16, 20, etc. and n=-4, -8, -12, etc., it will always give you 1 as the answer. Likewise n=1, 5, 9, 13, -3, -7, -11 will give i as the answer, and so on.

@PetruRatiu 3 жыл бұрын

Very cool video, but I feel it still missed a final step in the explanation at the end. Probably something like "exponents of the form 1/n are creating n different values that should be treated independently", but I'm still not entirely sure how to express that in a more rigorous form.

@t.minojan7029 2 жыл бұрын

While I was 12 years old I had a doubt -squre root(36) = -6 but I thinked differently if a * squreroot(b), we can write squreroot[(a^2) b] so -squre root(36) = squreroot[(-1)^2 * 36] then squreroot(1 *36) then squreroot(36) so answer is 6 wonderful memories

@artmaiq 5 ай бұрын

I love that in complex numbers, there always are n values that satisfy nth root

@Escviitash 3 жыл бұрын

Another wonderful example of argueing that you are just doing something, when you in fact are doing something else. Very similar to 0*2 = 0*1 -> 2 = 1 , where the argument is that you just CANCEL OUT a common factor, when you in reality DIVIDE BY a common factor. Well, the result will be the same if you can divide by the common factor, i.e. the common factor is anything but 0, infinity or negative infinty.

@grrgrrgrr0202 Жыл бұрын

The thing is that exponentiation is defined as x^a= exp(ln(x)*a). On complex numbers, the logarithm is not uniquely defined. And no matter what logarithm you take, the expression ln(exp(x))=x is not true for all complex numbers. However, on reals (using the usual logarithm that is only defined in positives), this expression was crucial for proving that (x^a)^b = x^(ab). For complex exponentiation, this expression just no longer holds.

@mr.unusual8509 7 ай бұрын

this is too complicated that i don't know what he's doing but i enjoy seeing new things haha

@MyNameIsSalo Жыл бұрын

essentially its the same as when we say sqrt(4) = 2 is incorrect, as it equals +2 and -2. Except the second the problem introduces imaginary numbers, then all imaginary and real roots need to be considered. Even the problem sqrt(4) = +/-2 is incorrect as there are imaginary roots, even though the problem doesn't mention imaginary numbers anywhere. If there is ever a fractional power, then all possible solutions need to be considered.

@nafrost2787 3 жыл бұрын

I found a similar paradox to this with real exponents: e^(2pi*i*ln(4))=(e^(2*pi*i))^ln(4)= 1^ln(4)=1 but according to Euler's formula, e^(2pi*i*ln(4)) doesn't equal 1. I would love if you can do a video about this with real exponents, because the explanation with gcd only works for rational ones at best.

@grrgrrgrr0202 Жыл бұрын

With real exponents, there are in fact countably infinitely many possibilities, depending on your choice of the complex logarithm.

@akshatrai7475 4 ай бұрын

If there are n roots of 1^1/n, then why do we say sqrt(1) is 1 only and not -1?

@rikschaaf 2 жыл бұрын

TLDR: the problem is that exponentiation is not a 1 to 1 function, but a n to 1 for the exponent being a whole number n and 1 to n for the exponent being 1/n where n is a whole number, so saying i^1 is equal to i^(4/4) turns the 1 to 1 function into an 4 to 1 to 4 function and you lose which goes to which if you treat it as i^4^(1/4) and turns the 1 to 1 function into an 1 to 4 to 1 function if you treat it as i^(1/4)^4.

@angryshavige1020 3 жыл бұрын

This was fun. :P

@binaryblade2 2 жыл бұрын

If you correctly treat i^4 = exp(i*2*pi*(4n+1)) which is always 1 but if you preserve it this way then exp(i*2*pi*(4n+1))^(1/4) = exp(i*pi*(4n+1)/2) which is the correct answer of i for all branches.