Largest open set for analytic function

Question

Give the largest open set where the function $f(z)=\sum_{n=1}^{\infty} ne^{nz}$ is analytic. Do I need to consider here the convergence of this Laurent series? I appreciate your kind help. Thank you so much!

Answer 1

Radius of Convergence of $\sum_{n=0}^{\infty}{nx^n}$ is $|x|<1$.

Hence, $|e^z|<1$ which is the half-plane $Re(z)<0$. (See Riemann Mapping Theorem)

Answer 2

The answer is $\{z\in \mathbb C: \Re z<0\}$. For the series to converge it is necessary that $ne^{nz} \to 0$. But $|e^{nz}|=e^{n\Re z}$ so it is necessary that $\Re z <0$. The series does converge whenever $\Re z <0$. Now we have to prove that it is analytic in $\{z\in \mathbb C: \Re z<0\}$. It is possible to prove differentiability directly using DCT but another approach is to prove that teh series converges uniformly on compact subsets of $\{z\in \mathbb C: \Re z<0\}$. If $K \subset \{z\in \mathbb C: \Re z<0\}$ is compact then there exists $\delta >0$ such that $\Re z <-\delta$ for all $z \in K$ Now just apply M-test to get uniform convergence.