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

[@official-obama]

NEVER calculate roots without a trusted adult's supervision

[@ItzMeBabbu]

I DO NOT give consent

[@NatoSkato]

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

[@нинажучкова-д2б]

but sqrt(1) is not equal to -1

[@NatoSkato]

@@нинажучкова-д2б then what is -1 squared

[@нинажучкова-д2б]

@@NatoSkato its not how it works my friend

[@Felixr2]

@@нинажучкова-д2б 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_]

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

[@blackpenredpen]

Good morning! Bprp, 4:55 am!

[@neilgerace355]

Good morning from 15 hours in the future!

[@kakashiuchiha1077]

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

[@jonatanpinsard5811]

Good moorning! Here is 9:10 am

[@8DJYash]

It's evening here

[@epsilia3611]

@@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)

[@yoavboaz1078]

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]

@@gtgagaggagagagga the step above implies that

[@DeJay7]

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]

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]

@@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]

@@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]

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

[@guitarninja0403]

Åland Islands 🇦🇽

[@ruffifuffler8711]

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]

@@ruffifuffler8711 What

[@ruffifuffler8711]

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

[@cougar2013]

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

[@tylershepard4269]

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]

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

[@Sam_on_YouTube]

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]

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

[@kiro9291]

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

[@RubyPiec]

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

[@Sam_on_YouTube]

@@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]

ah, yes, "Reapers"

[@lunstee]

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]

Yep I think you nailed it

[@damianbla4469]

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

[@analoghabits9217]

it's like wearing your face over your mask

[@watwat7050]

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

[@blackpenredpen]

Thanks!

[@stapler942]

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).

[@joaopedroalves1200]

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]

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]

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

[@createyourownfuture5410]

@@ThomasTheThermonuclearBomb thanks mate

[@wanpengli-mr8vv]

YES, the key is multivalued function

[@xXJ4FARGAMERXx]

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

[@SquibbyJ]

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]

Thank you!

[@Rofida-rh1wq]

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

[@HopeP0H]

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.

[@swisssr]

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]

Wowww tysm 🙇

[@AuroraNora3]

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]

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]

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]

@@Cobalt_Spirit arent x^n only principal roots?

[@Cobalt_Spirit]

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

[@gtgagaggagagagga]

@@Cobalt_Spirit eh of course *facepalms

[@joonjoon2368]

I’m just mesmerized by his marker quick switch

[@fghsgh]

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

[@rudrodeepchatterjee]

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.

[@angelmendez-rivera351]

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]

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]

@@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]

Thank you so much.❤️

[@XxfishpastexX]

thank you 😊

[@williamestey7294]

Well explained! Curious what do you do?

[@camrouxbg]

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.

[@bartekabuz855]

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

[@orionsrash1515]

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

[@UniqueNCS]

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

[@ryang628]

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]

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

[@pedrosso0]

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]

@@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]

(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

[@BryndanMeyerholtTheRealDeal]

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

[@RoderickEtheria]

Mistake is ruling out the complex roots when square rooting.

[@Viewer2812]

Yoo it's finally here.

[@aashsyed1277]

SAME FOR MEE

[@nicolasgoubin]

And here we are

[@akshatrai7475]

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

[@esajpsasipes2822]

i think i got it at about 4:20 - sqrt(x^2) is not x but rather |x|. Bet the same applies here - (for x^1/4)^4 the awnser is simply x and with 1 plugged in would be just 1, but for (x^4)^1/4 the awnser is |x| which is still 1 no matter which solution you put in as x

[@esajpsasipes2822]

well not quite correct but it still gets something right ig

[@miguelalejandromorenobarri4759]

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

[@eddyunterseher4430]

e^(2πni) = 1 | ln 2πni = 0 | /2πn i = 0/2πni i = 0 (n is an Integer)

[@aweebthatlovesmath4220]

2nπi=0 meaning that n is zero because 2 pi and i aren't when you divide by 2niπ what you actually were doing was which is undefined its undefined because it has infinitly many solutions. That's why i≠0

[@jeremysharpe5467]

-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).

[@angelmontoyagallardo5679]

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

[@paradoxreboot]

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

[@TitanOfClash]

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]

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]

@@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

[@cmilkau]

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.

[@ManuelRiccobono]

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

[@LuisBorja1981]

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

[@leftenanalim]

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

[@asparkdeity8717]

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

[@JG-ne2mx]

“Yes it was me 😁” great line

[@EvilDudeLOL]

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.

[@tri99er_]

Wait, was this just an elaborate way to establish, that exponentiation is not commutative?

[@manojsurya1005]

The conclusion at end was great

[@jithin.kesanapalli8317]

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

[@cmilkau]

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

[@finkployd6110]

Because there are four complex 4th-roots of 1: 1, -1, i, and -i.

[@MathElite]

Morning 8:30am EST Love your videos

[@akshatjangra4167]

Hello there!

[@SomeTomfoolery]

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

[@myidanny]

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]

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]

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

[@oenrn]

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.

[@cirax856]

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.

[@ulftho]

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.

[@ajshots748]

It’s like how x^2=25 makes x=-5,5 But x=sqrt(25) makes x=5

[@blackholedividedbyzero]

sqrt(25)=±5😊

[@ajshots748]

@@blackholedividedbyzero only when solving equations Some equations

[@lumina_]

@@blackholedividedbyzero no

[@Imran-Shah]

@@blackholedividedbyzero No, as an operator, √ only leads to the positive solution.

[@binaryblade2]

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.

[@ley2497]

A 5:50 I don't understand where come from the n and why it equals to 0,1,2,3

[@Imran-Shah]

DeMoivre

[@jibster5903]

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

[@hayn10]

i-th root of i multiplied by i^i = 1 pretty cool

[@sergeygaevoy6422]

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.

[@MrCigarro50]

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

[@refrainrestrainresist-3rs49]

Hello bprp! I am studying Complex Numbers and this video popped up. I liked your explanation! Could you also please explain why you've subtracted 2pi from the solutions at 12:36 ? My trigonometry is weak. How can I improve it? I have never been able to understand the radian calculations regarding domain, range etc. Edit: Is it because the sine and cosine function repeats after 2pi interval? Hence we are changing the angle while retaining the value. I heard your explanation once more and I believe I have understood. Please correct me if I'm wrong!

[@esajpsasipes2822]

you understand it correctly. You can freely substract or add the periode without changing the awnser, because the angle would be still the same, apart from full rotations. What he wanted is to get the awnser with the "base angle" - the angle that lies on

[@Escviitash]

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.

[@remo9229]

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

[@MyNameIsSalo]

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.

[@_kopcsi_]

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.

[@iamnotuta2658]

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

[@TomFarrell-p9z]

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.

[@johnernst2472]

How about delving into the intricacies of Mandelbrot fractals?

[@snekback.]

5:19 doesn't that rule not always apply for complex numbers? And isn't the proper way to do it this: z^w=e^(w*ln(z)) Knowing that e^(2ipi+1)=e will make this easier to read 😅 For example: [e^(2ipi+1)]^(2ipi+1)=e^[(2ipi+1)(ln{e^(2ipi+1)})] , however [e^(2ipi+1)]^(2ipi+1) != e^[(2ipi+1)(2ipi+1)]

[@sergeygaevoy6422]

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.

[@RuleofThehyperbolic]

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

[@enricofermi5608]

We have (Euler's identity): e^(2i π)=1 e^(i π)= -1 So: -1= e^(i π)= e^[2i π*(1/2)] = [e^(2i π)]^(1/2) =1^(1/2) =1 and finaly a CONTRADICTION: -1=1 Where is the error ?

[@Andyg2g]

What confuses me is that we are taught that expressions of the form a^b^c are to be evaluated as a^(b^c) in a kind of “top-down” approach. Is there a simple way to understand the difference here?

[@sylv512]

the beard makes your maths 10x better

[@guillaume5313]

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

[@Alisayadi2000]

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❤️

[@moreanushka4701]

2:57 but will it means that 1 is equal to all four roots. Because any root of 1 is always equals 1 except infinity♾️

[@davidwright8432]

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

[@nafrost2787]

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]

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

[@romaobraz4295]

this is brilliant. great vid!

[@alexandrentema6202]

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

[@kiro9291]

that was a satisfying explanation, thank you

[@glennjohnson4919]

What is i to the 1/pi? If you use the polar decomposition, you get any unimodular complex number. Do “powers” of complex numbers really make sense? There are solutions to polynomial equations, and multiplication of complex numbers is defined, but the laws of exponentiation are trouble without clarification.

[@grrgrrgrr0202]

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.

[@dennisrosero]

Hi, i have a question, it means that sqrt(1) has 2 solutions? I think the operation only have one solution, indeed fourth root of 1

[@blackpenredpen]

I think it really depends on the notation and the context. Usually, I will give only 1 answer when we compute √(1), which is 1 because the symbol √ is defined to be the principal square root. But if the question is written in words "find ALL the square roots of 1" then I will give 2 answers, 1 and -1.

[@danelrosen5461]

There is two ways of see sqrt(x) The funtion form Sqrt(x)=y Where for each value of x corresponds only one value of y And the algebraic form Sqrt(x) = |y| Where for each value of x corresponds two values of y

[@TheJaguar1983]

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 😂

[@JayTemple]

This has little to do with the problem, but with nomenclature. What do you tell students about reading the expression "-a"? There are teachers who say you absolutely should not call it "negative a," because a might be positive. There are also teachers who say you shouldn't read it as "minus a," although I forget their reasoning. (I tell my own students that both are fine. "Negative a" is negative 1 times a, and "minus a" is the result when you take 0 - a.) The reason I'm curious is that you referred to -i as "negative i," and you undoubtedly share my experience of demonstrating why complex numbers do not have signs in the sense of the real numbers.

[@meatystalactite531]

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

[@Setiny]

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

[@arnavoza527]

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

[@pronounjow]

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]

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

[@pronounjow]

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.

[@Bikerider919-23]

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

[@nicolasgoubin]

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

[@frogstarian]

This is like a phenomenon I came up with in college. -1=-1 -> -1=1/-1 -> sqrt(-1)= sqrt(1/-1) = sqrt(1)/sqrt(-1) -> i = 1/i -> i=-i -> 1=-1 The problem is fixed if we don't allow negatives to be in the denominator when taking even roots, but it really threw me for a loop for a while.

[@max_trnw]

1:09 who would even think about voting Step 1

[@cougar2013]

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

[@TheBrainReal]

I've been wondering, what is the derrivative of x^^x? Is it even possible to calculate?

[@autumn_auburn]

Did you mean to say x^x ? If yes, then its derivative is (x^x)(1+ln(x)). You can find it by taking logarithm of y=x^x and then differentiating both sides.

[@TheBrainReal]

@@autumn_auburn no I meant the power tower of x ^ x ^ x ^ x ... with height x. X double up arrow X

[@oledakaajel]

We don't have a continuous version of tetration so no

[@angelmendez-rivera351]

x^^x is not well-defined for arbitrary real x, so it is not possible to evaluate x^^x, let alone take its derivative. At the very best, you can let x be a half integer greater than -2, and it will work, but nothing more than that.

[@autumn_auburn]

@@TheBrainReal Oh, I see. Thanks, now I am curious too!

[@danielschwartz6795]

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

[@StefanReich]

0:46 Ryyyyye?

[@PastaSenpai]

bruh 💀

[@StefanReich]

@@PastaSenpai lol

[@oscarmartinpico5369]

My question is how to control the root of any power to choose the one that applies through the development of the operations. Even, with the root square I have troubles.

[@t.minojan7029]

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

[@thatrandomharpguy7564]

e^iπ/4 is also a fourth root of 1 :D