Convergence of complex power series question

Question

I need some help to solve this problem and find the domain of convergence of the following power series:

$$\displaystyle\sum_{n=0}^\infty(2^n+i^n)(z-2i)^n$$

Thank you!

Answer 1

Let $a_n=2^n+i^n$. The radius of convergence of the entiere serie: $\sum a_n z^n$ is $R=\frac 12$ since $$\lim_{n \to +\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to +\infty}\left|\frac{2+\left(\frac{i}{2}\right)^{n+1}}{1+\frac 12\left(\frac{i}{2}\right)^{n}}\right|=2$$

Then your serie is absolutely convregent if $|z-2i| < \frac 12$ and divergente if $|z-2i| > \frac 12$

Answer 2

Note that $$ \lvert 2^n+i^n\rvert=\sqrt{2^{2n}+1}, $$ and hence $$ \lvert 2^n+i^n\rvert^{1/n}=(2^{2n}+1)^{1/2n}, $$ and $$ 2<(2^{2n}+1)^{1/2n}<(2^{2n}+2^{2n})^{1/2n}=2\cdot 2^{1/2n}\to 2, $$ which implies (root test) that the radius of convergence of the series in $$ r=\frac{1}{2}. $$