Which complex numbers satisfy $(-1)^z=1$ and $(-1)^z=-1$?

Question

I tried to generalize the notion of parity towards complex numbers but Wolfram Alpha refuses to help me.

UPDATE

I meant separate equations, not a system of equations.

Answer 1

You want to find solutions to the two (separate) equations:

$(-1)^z = 1$

and $(-1)^z = -1$.

Let's tackle the first. Taking something to a complex power is complicated as the result can have multiple, even infinite, distinct values. So we need to be precise about what we want. So I'm going to interpret the problem as: "find all $z \in \mathbb{C}$ such that at least one possible value of $(-1)^z$ will be equal to $1$".

When you're solving a purely real value problem like $a^x = b$, you take logs of both sides. You can do the same here with the complex valued logarithm. However, unlike the real logarithm, it can take an infinite number of values and you must consider the general form.

So: $\log (-1)^z = \log 1$

$z\log(-1) = \log 1$

Noting that $-1 = e^{i\pi(2n+1)}, n \in \mathbb{Z}$ and $1 = e^{i\pi 2m}, m \in \mathbb{Z}$ allow us to compute the general form of the complex logs and write:

$zi\pi(2n+1) = i\pi 2m$

$z = \frac{2m}{2n+1}, \ \ m,n \in \mathbb{Z}$

and that's the general solution for the first equation. Note that, setting different integer values (including $0$) for $m,n$, we can generate all the solutions, including $z = 0, 2, -2, \frac 25, -\frac 23$ et cetera.

You can handle the second equation the same way, with the general solution here being $z = \frac{2m+1}{2n+1}, \ \ m,n \in \mathbb{Z}$.

Specific solutions include $z=1, -1, \frac 35, -\frac 37$ and so on and so forth.

Note that the solutions to each equation are all rational real numbers.

Answer 2

There is no $z$ such that $(-1)^z=1$ and $(-1)^z=-1$.

Proof : assume there is a number $z$ such that $(-1)^z=1$ and $(-1)^z=-1$. then $(-1)^z=1=(-1)^z=-1$

or $1=-1$. contradiction QED.