0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $|e^{-ib\ln n}|=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.