Number of roots of a function

Question

If $z^3= 2$, one has 3 roots.

If $z^4= 2$, one has 4 roots.

What about $z^\pi = 2$ ? How many roots it has? And what is the procedure to find them?

Answer 1

Serious answer, assuming you are familiar with Euler's identity, etc, let $$z=re^{i\theta}\implies z^\pi=r^\pi e^{\pi i\theta},$$ Therefore $r=2^{\frac{1}{\pi}},$ and $\pi\theta=2\pi k$ for any $k\in\mathbb{Z}$; hence $$\theta=2k.$$ Or in cartesian coordinates, $$z=2^\frac{1}{\pi}(\cos(2k)+i\sin(2k)).$$ Due to these angles being irrational multiples of $\pi$, they are countably infinite in number.

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Small addendum: It's usually stated that the exponent tells us how many roots a complex number has, but if we want to be more technical, it's something of a measure of the rotational symmetry of the roots. There are infinite roots for all complex numbers, but for integer (and rational) exponents, these roots perfectly overlap, producing only finitely many distinct roots.