Where $\log(1+z^2)$ is complex differentiable? Where $\log(1+z^2)$ is complex differentiable? 
I use the fact, that logarithm of $1+z$ is equal to its Taylor series expansion whenever $|z| \le 1$ and $z\ne -1$.
So the answer is that $\log(1+z^2)$ is complex differentaible when $|z|\le1$ and $z\ne i, z\ne-i$.
Is that correct?
I'm not sure if that is enough.

We first find the inverse image of the negative real axis (and $0$) for $g(z)=1+z^2$.This is found to be the part of the imaginary axis with modulus $\geq 1$. $f(z)=log (1+z^2)$ is defined outside this set and differentiable there by the chain rule giving $f'(z)=\frac{2z}{1+z^2} $.
