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

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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[official-obama]

NEVER calculate roots without a trusted adult's supervision

[NatoSkato]

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

[yoavboaz]

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

[vibby]

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

[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.

[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.

[tylershepard]

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.

[watwat]

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

[stapler]

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).

[damianbla]

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

[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

[joaopedroalves]

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.

[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

[HopePH]

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.

[joonjoon]

I’m just mesmerized by his marker quick switch

[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.

[orionsrash]

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

[angelmendez-rivera]

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.

[UniqueNCS]

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

[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!