Find the radius of converges $$\sum_{n=o}^{\infty}\frac{(z-1+i)^n}{2^n+n} $$
Let $w=z-1+i$
$$\sum_{n=o}^{\infty}\frac{w^n}{2^n+n} $$
$$R=lim_{n \to \infty}|\frac{a_n}{a_{n+1}}|=lim_{n \to \infty}\frac{2^{(n+1)}+n+1}{2^n+n}=2$$
So there is series converge for $|w|<2$
for $|w|=2$ we need to look at $$\sum_{n=o}^{\infty}|\frac{w^n}{2^n+n}|=\sum_{n=o}^{\infty}\frac{|w|^n}{|2^n+n|}=\sum_{n=o}^{\infty}\frac{2^n}{2^n+n}$$
But $$\lim_{n\to \infty}\frac{2^n}{2^n+n}=1$$ so it does not converge.
$$|w|<2\iff |z-1+i|<2\iff (x-1)^2+(y+1)^2<4$$
So the series convarges at the interior of the circle around $(1,-1)$ or radius $2$
Is this correct?
Remember the ratio test says the series converges if
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$