Convergence of $\sum_{n=1}^{\infty} \frac{(\cos(iz))^n}{n} $ 
Examine the series convergence : $$\sum_{n=1}^{\infty} \frac{(\cos(iz))^n}{n}$$

I transformed it into $$\sum_{n=1}^{\infty} \frac{(e^z+e^{-z})^n}{n2^n}$$ and I don't know what to do now.
 A: Let $$S(z)=\sum_{n\ge1}\frac{\cos(iz)^n}{n}.$$
Note that $$\frac1{1-x}=\sum_{n\ge0}x^n,\qquad |x|<1,$$
so
$$-\ln(1-x)=\sum_{n\ge1}\frac{x^n}{n},\qquad |x|\le 1, x\ne 1.$$
If $z\in\Bbb R$, $\cos(iz)=\cosh(z)$, and the series $S(z)$ diverges.
Let $z=a+bi$. We may see that
$$\cos(i(a+bi))=\cosh(a)\cos(b)+i\sinh(a)\sin(b).$$
Thus
$$\cos(iz)=re^{iT(z)},$$
where
$$r=\sqrt{\sinh^2(a)+\cos^2(b)},$$
and $$T(z)=T\in\left\{\arcsin\left(\frac{\sinh(a)\sin(b)}{r}\right),\quad \arccos\left(\frac{\cosh(a)\cos(b)}{r}\right),\quad \arctan\left(\tanh(a)\tan(b)\right)\right\}.$$
We know that regardless of the choice of $t$, $|e^{it}|=1$.
We can split the rest up into cases.
Case 1: $r< 1$ and $T(z)/\pi\not\in\Bbb Z$.
In this case we have that $|z|=|re^{iT}|=|r|<1$ so $S(z)$ converges.
Case 2: $r=1$ and $T(z)/\pi\not\in\Bbb Z$.
In this case $|z|=|re^{iT}|=|e^{iT}|=1$. But since $T/\pi$ is not an integer, $e^{iT}\ne 1$, so $S(z)$ converges to $-\ln(1-e^{iT(z)})$.
Case 3: $r>1$ and $T(z)/\pi\not\in\Bbb Z$.
This implies $|z|>1$ so $S(z)$ diverges.
Case 4: $r<1$ and $T(z)/\pi\in\Bbb Z$.
This gives $|z|=|r|<1$ so $S(z)$ converges.
Case 5: $r=1$ and $t(z)/\pi\in\Bbb Z$.
This gives $z=(-1)^{T/\pi}$. If $T/\pi$ is even, $z=1$ and $S(z)$ diverges. If $T/\pi$ is odd, $z=-1$ and $S(z)$ converges to $-\ln 2$.
Case 6: $r>1$ and $T/\pi\in\Bbb Z$.
This gives $|z|=|r|>1$ so $S(z)$ diverges.

To sum up (pun very intended), the series $S(z)$ converges iff $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}<1,$$ OR $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}=1\quad \text{AND}\quad \frac{1}{\pi}T(z)\not\in\Bbb Z,$$ OR $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}=1\quad \text{AND}\quad \frac{1}{\pi}T(z)\in\Bbb Z \text{  is odd}.$$

