What is (-1)^π equal to?

Just a short problem for the upcoming Pi Day! Can you figure it out?

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[pit]

Good video, because it raises two big problems about raising a number to a power, and you quite fell into the trap. The two problems are: is the number (-1)^π even defined, and are the calculations you've done correct? The quick answer to both of these questions is "well yes, but actually no".
I'll try to do my best to go into the details.


First, how raising a number to a power is defined? Let x and y be two numbers. If x is any complex number, and y is an integer, then x^y is defined as the product of x times itself y times. For example, (-1)^2 = (-1)*(-1) = 1.
Now, if y is any real number, x has to be positive for x^y to exist: x^y is defined as exp(y*log(x)), where exp is the real exponential function, and log the logarithm function, i.e. the reciprocal function of exp. Here log(x) exists only if x>0, which is why x has to be positive.
Then, using the extension of exp to the complex plane, x^y can be defined when y is any real number, and x still has to be a positive real number. The definition of x^y is the same (i.e. x^y = exp(y*log(x))), except that exp is now the complex exponential function.
So here, we've made quite a progress to define x^y as much as we can, and (-1)^π is still not defined. If we want to go further, we would want to consider the extension of log to the whole complex plane, so that x^y = exp(y*log(x)) can be defined with x not being a positive real number (recall the we want x to be -1). However, this is where we get in trouble, because such a thing doesn't really exist. What's the problem here? We want to define the reciprocal function of exp (the complex exponential function), i.e., we want to define log(z), with z a complex number, as the solution u of the equation exp(u) = z. If we have a solution u, we can set log(z) = u. But if you try to do so, you'll notice that there are infinitely many solutions for u, and thus log(z) has infinitely many values. It happens because exp is periodic: exp(z+2iπ) = exp(z) * exp(2iπ) = exp(z) * 1 = exp(z). Notice however that exp(u) = 0 has no solution, so log(0) will still be undefined.
Here, we have two solutions:
- Either we are fine with multivalued functions, thus x^y is also multivalued, and the computation of (-1)^π has infinitely many answers, one of them being the one you're given in your video.
- Or you prefer functions that have only one output. If so, you just have to pick one output for each input, i.e. if you want to define log(z), solve the equation exp(u) = z for u, find infinitely many solutions, and pick the one you want. However, if you do this in a random way, you'll end up with a logarithm function that won't be very nice (i.e. continuous and derivable in the complex sens). To make log nice, you have to pick all the u's in a nice way. The common way to do so is to pick u such that its imaginary part is in the open interval (-π, π). However, doing so implies that log is not defined for complex numbers z of the form z = r*exp(-π), with r a positive real number, and especially for -1. So log(-1) is still not defined. This can be circumvent by taking another open interval. And here is the problem: there is a as many complex logarithm functions as open intervals of length 2π. So if you take the open interval to be (0, 2π), (-1)^π will be defined and it will have a certain value, but if you pick another open interval, let's say (-2π, 0), (-1)^π will still be defined, but it will have a different value.
So all in all, (-1)^π can be defined, but it's definition (and thus its value) depends on the exponential logarithm you consider.

Secondly, you're computation is kind of false: (x^y)^z is not always equal to x^(y*z). There is some cases when it's actually true, for example when x, y and z are all positive real numbers, or when x is positive, y is complex and z is an integer. But when x is negative or z is not an integer, (x^y)^z can be different from to x^(y*z). For example, with x = e, y = 2iπ and z = 1/2, you have (x^y)^z = 1^(1/2) = 1 but x^(y*z) = e^(iπ) = -1. You have an even simpler example with x = -1, y = 2 and z = 1/2 (this is in fact the same idea, but expressed only with integers and rational numbers).

What I want to point out with my comment is that what you're showing in your video is not actually true, even if it can be true in some sens (and all this notion of trueness heavily depends on the hardness of defining a good complex logarithm).

[tokajileo]

strange that manipulating two real numbers results in a complex number.

[danmerget]

Hang on. That answer seems incomplete to me. There should be multiple answers, because there are multiple values of theta such that e**(i*theta) = -1. It's not only true that e**(i*pi) = -1, but it's also that e**(3i*pi) = -1, as well as e**(5i*pi), e**(7i*pi), etc. In general, you can add any multiple of 2*pi to theta.

So I think the actual answer should be cos((2n+1) * pi**2) + i * sin((2n+1) * pi**2), for any integer value of n. And please forgive the crude ASCII calculations.

[Bibibosh]

these videos help me to continue being inspired by mathematical problems!

[adityaghosh]

I think (-1)^pi has multiple values. Because instead of e^ipi, you could have put e^(i(2n+1)pi) which yields a set of values for (-1)^pi.

[NestorAbad]

It's worth noting that proceeding the same way we can easily prove that i^i is... a real number!
Thanks for sharing!

[MrKA]

sir, you're wrong.
-1=cos(pi+2*k*pi)+i where k= ..-2, -1, 0, 1, 2, .. is an integer. So you have an infinite number of values for (-1)^pi.
What you calculated is only the so called principal value of -1^pi, i.e for k=0.

[giantsfan]

Euler's formula, greatest of all time.

[asfermallick]

Sir, your videos are splendid.
I love watching them.I enjoy the problems on geometry.Make more of it

[prakharrai]

Some people say that your videos are I think it is all about the But really appreciating that you cover all parts :EASY TO HARDEST MATH PROBLEMS

[username]

For anyone wondering, in the reals, powers with irrational exponents are ONLY defined for positive bases. Furthermore, as others have pointed out, the multiplicative power rule is false in general in C.

[LogicalMath]

AB * A * B = BBB, where AB is a 2 digit number, and BBB is a 3 digit number. Find A and B.

[BriTheMathGuy]

Very cool. Reminds me of how i^i is a real number.

[math_travel]

it is very interesting problem I haven't thought of. It feels very strange that the value is set as one. thank you for inspiring me!

[Kllazik]

I used complex e^(ln(-1)^pi) then I put pi in front of ln and solved complex logarithm which is equal to ln1 + ipi, so so in the end I also got the same result e^ipi^2. BTW Nice video :)

[seanfraser]

I know that you didn’t want to make this video too confusing, but you weren’t careful about your choice of logarithm of -1. You chose i*pi, which is valid, but if you were being careful you would use (2k+1)*pi*i.

Thus
(-1)^pi = cos((2k+1)pi^2) + i*sin((2k+1)pi^2)

is the most correct solution.

[mohd.talhaansari]

Hey! I had a question
How can we locate 'pie' on a number line ?

[Ariana-dnmm]

Note that this actually has infinite solutions since
-1=e^(i(π+2nπ))
So solutions are

where n is a integrr

[ishusingh]

Why we cant write it as

= [(-1)^1/7]^22
= [-1]^22
= 1

[scarstreet]

Hi, I got a (perhaps easy) question. Isn't π = 22/7? If not, then what π is this question referring to and also what's 22/7? And if π is 22/7, why can't we just use that?