Complex powers in $\mathbb{R}$

Question

Problem: For which of the following complex numbers $z\in\mathbb{C}$ is $ z^8\in \mathbb{R}$?

a) $z=-5+5i$
b) $z=100+i$

Of course, one can go the hard way and apply the binomial theorem or compute everything by hand, which will give us the following:

a) $z^8 = 6250000\;$ and
b) $z^8=9.972007 \cdot 10^{15} + 7.99440056 \cdot 10^{14} i$

Question: Is there a faster way to find out that $z\in\mathbb{R}$ for a) and $z\in\mathbb{C}\setminus\mathbb{R}$ for b)?

Answer 1

For part $(a)$: $$z= -5+5i = -5(1-i)=-5\sqrt 2 e^{-\pi /4 i}$$

$$z^8 = (-5\sqrt 2)^8 e^{-2\pi i}=625000$$

For part $(b)$:

$$z=100+i = \sqrt {10001} e^{\alpha i}$$ where $$\alpha = \cos^{-1} \frac {100}{\sqrt {10001}}$$

$$z^8=10001^4 e^{8\alpha} = 10001(\cos 8\alpha + i\sin 8\alpha)$$

Answer 2

you could use $arg(z^n)=n\cdot arg(z)$.

In case a) you have $arg(z)=2\pi\cdot 7/8$; thus $arg(z^8)=2\pi\cdot 7$, which clearly is on the real axis.

--- rk