composition of complex functions

Question

I'm given that $g(z)= \ln r+i\theta$ where $(r>0,0<\theta<2\pi)$ . I've already shown this function is analytic and that its derivative is $g'(z)=\frac{1}{z}$. Now it wants me to show $G(z)=g(z^2+1)$ is analytic where $x,y>0$ but what I don't get is what the output of $g(z^2+1)$ looks like. I know that $z=re^{i\theta}\implies z^2=r^2e^{i2\theta}\implies z^2+1=r^2e^{i2\theta}+1$. Then $g(z^2+1)=g(r^2e^{i2\theta}+1)$? From here I just dont see what the output would be. Is it $g(r^2e^{i2\theta}+1)=\ln r^2+i2\theta$?

Answer 1

Yes, $g(z^2+1) = g(r^2e^{2i\theta}+1)$.

But, $g(z^2+1)$ is more complicated than $$\ln(r^2) + 2i\theta$$ (which is $2g(z)$ btw), you have forgoten the "+1".

I don't think this output is relevant: let $H^+=\{z\in \mathbb C ; y>0, x>0\}$ and $\Omega=\mathbb C \setminus \{z\in \mathbb C ; \theta \equiv 0 [2pi]\}$. So $$f:H^+ \to \Omega, z\mapsto z^2+1$$ and $$g:\Omega \to \mathbb C$$ are both analytics.

So the composed function $g o f$ is analytic. You just have to show properly that $f(H^+) \subset \Omega$.