# Complex Number Upper Estimation

Hello I am trying to understand the following solution but cannot

Let $$a_{n}$$ be a bounded sequence of complex numbers. Show that for each $$\varepsilon>0$$, the series $$\sum_{n=1}^{\infty} a_{n} n^{-z}$$ converges uniformly for $$\operatorname{Re} z \geq 1+\varepsilon$$. Here we choose the principal branch of $$n^{-z}$$.

Solution $$\sum_{n=1}^{\infty} a_{n} n^{-z},\left|a_{n}\right| \leqslant C, \operatorname{Re} z \geqslant 1+\varepsilon .$$ Apply Weierstrass M-test, $$\left|a_{n} n^{-z}\right| \leqslant$$ $$C n^{-\operatorname{Re} z} \leqslant \frac{C}{n^{1+\varepsilon}}=M_{n}, \sum M_{n}<\infty \Rightarrow \sum a_{n} n^{-z}$$ converges uniformly for $$\operatorname{Re} z \geqslant 1+\varepsilon$$.

Specifically on the line where it says $$\left|a_{n} n^{-z}\right| \leqslant$$ $$C n^{-\operatorname{Re} z}$$. I understand that $$|a_n|$$ is bounded by $$C$$ but why is $$|n^{-z}| = n^{-\Re(z)}$$? Which properties of complex number does this come from?

Just write $$z=a+ib$$ with $$(a,b)\in\mathbb R^2$$. Then $$|n^{-z}|=|n^{-a-ib}|=|n^{-a}||e^{-ib\ln n}|=n^{-a}$$ since $$|e^{-ib\ln n}|=1$$.