Calculating i to the power i the right way. Why every proof you have seen is wrong

0:00 calculation
2:11 complex numbers
3:30 "proof" 1
6:21 "proof" 2
8:20 complex exp
9:12 complex log
12:32 the right way

WolframAlpha
Math StackExchange
Complex Variables and Applications, James Ward Brown, Ruel V. Churchill, 7th edition, Chapter 1 Section 8, Roots of Complex Numbers
The complex logarithm, exponential and power functions, Professor Howard Haber

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[pelegsap]

I really like the canceling digits method example 😂

[dekeltal]

(a^b)^c is indeed not always applicable to complex numbers, however, your example is *wrong*.
1 = e^2PI*i
then you raise both sides to the power of 1/2, effectively taking the square root. The square root function is not injective, you should have added +- to one of the sides, which would have made the equation correct.
Basically you've shown that both 1 and -1 are square roots of 1 (which is not a mistake)

[martind]

There is nothing wrong with the first two proofs.
Those are functions that give multiple branches, which is why you can produce an apparent contradiction, but in each case the calculation does still give _A_ branch, and that branch is a valid evaluation.

[kicorse]

At 6:12, the reason you get a nonsense answer here is independent of the (a^b)^c = a^(bc) step. You've stated that 1^(1/2) = 1, which is correct, used (a^b)^c = a^(bc) to find that 1^(1/2) = -1, which is also correct. The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another.

If you are careful to always note that e^f(x) = e^(f(x) + 2 i n pi), where n is an integer, then that avoids the problems associated with (a^b)^c = a^(bc). Of course, using this method isn't helpful when the exponent has a base of 1, because using a base of 1 is generally problematic and breaks a lot of general rules. For example, 1^(infinity) is undefined rather than infinite.

[gooseminecraft]

with complex numbers, the log function and fraction exponents are multivalued. in the steps where you "disproved" rules of math for complex numbers you conveniently decided to ignore the multivalued nature of these operations, which you can not do.

[sylverfyre]

Your counterexample to proof 1 is finding out that 1 has two square roots, one of which is 1, and the other of which is -1. This is not wrong. That's always something you need to consider when undoing a root with a square! You're focusing on an extraneous root.

[pauselab]

it depends on how you define complex logarithms. if you want to have a function, you need to allow a principal value of the logarithm. if you want to find all solutions, it isn't a normal function anymore because it has multiple outputs. method 2 isn't a "wrong" proof, it just glosses over some small detail with +2*k*pi*i in the logarithm function which does cancel out when you raise it to e. It's obviously different from the fraction coincidence at the beginning because this is a good method that always works for any similar problem. you can proof euler's theorem with taylor series or you could just assume the result. doesn't make it wrong.

[Ocklepod]

doesnt the 1=-1 proof fail because you decided that
squareroot 1 = 1
instead of
squareroot 1 = +-1 ?
For complex numbers, if you account for all branches the identity (a^b)^c=a^(bc) should hold.

[JaybeePenaflor]

I love complex analysis! In the case of the natural logarithmic function, we can make a branch cut and define the principal branch as log z = ln r + iθ, where r > 0 and θ is in (-π, π]..

[migssdz]

I mean, the second proof was basicaly showing that e^(-pi/2) is an answer. The only thing it did differently from the final proof was considering just 1 answer to ln(x), but this answer is still valid so it should also get only one answer to i^i (and it did)

[JakubS]

Isn't this the method that Blackpenredpen uses? So you wouldn't be the only one that uses a justified method.

[FalseNoob]

Dam this definitely blew my mind🤯
I have been so used to doing these problems normally that I almost forgot the limits of the basic mathematics rules like this indices one!

[killallbots]

In 4:43 The identity *_(a^b)^c = a^(bc)_* is not only true with *_b_* and *_c_* as real numbers. It is also true as long as the variable *_a_* is a positive real number and that the variable *_b_* is in the form *_Log z_* or is multivalued with respect to *_2πni_* where *_n_* is any integer. The variable *_c_* could be any non-multivalued complex number. This is proven in 8:56 where Presh showed that *_z^w = (e^Ln z)^w = e^(w Ln z)_*

[monoamiga]

Than explains so much! Thank you! No matter if it's i^i or whatever else, the fact you made me pay more attention while carrying out supposedly obvious operations is priceless.

[juxx]

Presh, you're usually right, but this time... All the values of the "counter-proofs" are a bit of fraudulent.

In the first, you took not the principal value of one in the complex plane but another one (1=e^2πi) and you stated that is equal to one (true). But then you took the square root. 1 is the principal complex square root of the complex number 1, but you took the complex square root (which is multi-valued) of e^2πi. In other terms, you took the real square root on the LHS (which matches with the principal complex square root on complex numbers with imaginary part = 0) and the complex square root (multivalued) in RHS. If you treated 1 with the general argument (e^i(0+2πk), k is an integer) you wouldn't get to this result because it has another value which is 1.

In the other, happens the same thing. You took the real natural logarithm on the LHS (which matches the principal branch of the natural logarithm on complex numbers with imaginary part = 0) and the complex natural logarithm on the RHS (the multi-valued natural logarithm). That property has to be true, not also because you use it when solving the equation z = e^w, but because is the definition of the natural logarithm (and it has to be multi-valued because so it is the complex exponential)

In both cases, you took two different functions on the two sides, making the answer blatantly wrong. In complex numbers, is better to work with the general argument and then reduce it to the principal form if you're not sure what you're doing.

Technical misunderstandings: Ln(z) is the notation of principal branch of natural logarithm, usually not the notation for real natural logarithm. It's subtle because for real numbers is equal, but not for complex numbers. Use Log(z) and log(z) notation for complex natural logarithm (since defining based-logarithms is a bit useless in the complex world) and ln(x) for natural real logarithm.

Thank you for reading to anyone who kept reading so far and I love you Presh, this is just constructive criticism

[CorruptMem]

8:00 The natural log of any number except zero has infinite solutions. When you take the ln of 1, it would give the following solutions: ln(1) = 0 + n*2*pi*i, where n is an integer. Similar to ln(e^(2*pi*i)), it would result in: ln(e^2*pi*i) = 2*pi*i + n*2*pi*i = (n+1)*2*pi*i = m*2*pi*i, where m is an integer. This means: ln(1) = ln(e^(2*pi*i)).

[BakersTuts]

Wait so what happened to the infinite series at 9:00? Why show that if it wasn’t used?

[enantiodromia]

When, in "proof" 1, you substitute 1^(1/2) for 1, you must also say that 1^(1/2) has two results, namely 1 and - 1. So the power law holds for one of the values.

[donplay]

you have taken 2nd principal value for 1 which will give you -1 its no surprise there as 1^1/2 may equal to -1. but for 1=e^0 1st principal value it will remain 1. Well new method is also correct but the previous method also works we just need to be careful. as we have applied the the method to find roots of unity x^5=1.

[karhukivi]

This has to be one of your best ever videos, if not the best - bravo!