In what area the function $\ln(1+z^2)$ is not analytic?

Question

If $\ln(z)$ is the main branch of the the logarithm ($-\pi<\arg z<\pi$) then in what area the function $\ln(1+z^2)$ is not analytic ($z$ is a complex number)?

The method to solve this problem as I guess is to separate the real and imaginary parts and use the Cauchy-Riemann equation. then the areas in which the equation doesn't hold is the answer. I'm not sure if my method is right, even if it is, I don't know how to separate correctly and get the right result.

Introducing books or websites which explain this topic will be very helpful.

thanks in advance.

Answer 1

Hint:

$$\log(1+z^2)=\log(1+x^2-y^2+2ixy)=\ln\sqrt{(1+x^2-y^2)^2+(2xy)^2}+i\arctan\dfrac{2xy}{1+x^2-y^2}$$ then work with $$u=\dfrac12\ln\left((1+x^2-y^2)^2+(2xy)^2\right)~~~,~~~v=\arctan\dfrac{2xy}{1+x^2-y^2}$$

Answer 2

Since you are defining $\ln$ as the main branch of the logarithm, $\ln$ is an analytic function. And since the composition of analytic functions is again an analytic function, your function is analytic. More precisely, if$$D=\mathbb{C}\setminus\bigl\{z\in\mathbb{C}\,|\,1+z^2\notin(-\infty,0]\bigr\},$$then$$\begin{array}{ccc}D&\longrightarrow&\mathbb C\\z&\mapsto&\ln(1+z^2)\end{array}$$is an analytic function.