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.
 A: 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} $.
A: Let $z + i = r_1e^{i\theta_1}$ and $z - i = r_2e^{i\theta_2}$. Now
$$
\log(z^2+1) = \log[(z - i)(z + i)] = \log(r_1r_2e^{i(\theta_1 + \theta_2)}) = \log\lvert r_1r_2\rvert + i[\theta_1 + \theta_2]
$$
If we go $2\pi$ around $z + i$, then we have $r_1e^{i(\theta_1 + 2\pi)}$ which leads to
$$
\log(z^2+1) = \log[(z - i)(z + i)] = \log(r_1r_2e^{i(\theta_1 + 2\pi + \theta_2)}) = \log\lvert r_1r_2\rvert + i[\theta_1 + \theta_2] + 2\pi i
$$
We change by $2\pi$ so it isn't single valued. Same occurs if we go around $z = -i$. If we encircle both, we change by $4\pi i$. Additionally, $z=\infty$ is a branch point. To stop the encircling of the branch points we draw branch cuts to make the $f(z)$ analytic. That is, the branch cuts of $(-\infty, -i]\cup [i, \infty)$ will suffice. To prove that $z=\infty$ is a branch point, let $z = 1/\zeta$. Then $f(z) = \log\bigl(\frac{1 + \zeta^2}{\zeta^2}\bigr) = \log(1 + \zeta^2) - 2\log(\zeta)$ so $\zeta = 0$ is branch point then $z = \infty$ is one.
So $f$ is analytic in $\mathbb{C} - (-\infty, -i]\cup [i, \infty)$.
