Find the minimum $n$ such that $z^n\in(-\infty,0)$

Question

Given the complex number $z=$ $1\over i$ $+$ $1\over 1-i$ I have to find the minimum $n$ such that $z^n\in(-\infty,0)$. After a few simplifications I came up with the algebraic form $z=\frac12$$-$$\frac12i$, the modulus $\lvert z\rvert = $$1\over\sqrt2$ and the argument $\theta=-\frac\pi4$. How should I proceed?

Answer 1

Since the argument of $z$ is $-\frac{\pi}{4}$, the argument of $z^n$ will be $-\frac{n\pi}{4}$, up to addition by integer multiples of $2\pi$.

Note that $z^n\in(-\infty,0)\iff \arg(z^n)=\pi$, so we are finding the smallest integer $n$ such that $\pi$ and $-\frac{n\pi}{4}$ differ by a multiple of $2\pi$. You can check and see that this first occurs when $n=4$.