Branch point of $\log(z)+arcsin(z)$

Question

i try to do it by substituting $z$ into $\frac{1}{t}$. Then, it would become this function $-iln[i+t\sqrt{(1-\frac{1}{t^2})}]$. Teacher says that when $z=z_\infty$ is a branch point, only could be that $f(z)$ isn't finite. However, i couldn't figure out.

Answer 1

I assume you mean $i \ln z + \arcsin z$. As a multi-valued function, $$i \ln z + \arcsin z = \frac \pi 2 - i \ln \left( 1 + \sqrt {1 - \frac 1 {z^2}} \right).$$ From this it can be seen that there is an infinite number of sheets which are unbranched over $\infty$ (corresponding to the positive square root) but also an infinite number of sheets which are sequentially glued over $\infty$ (corresponding to the negative square root). There is still a logarithmic branch point at $\infty$ if you add any multiple of $\ln z$ to $\arcsin z$.

Finiteness isn't really relevant here; $1/\sqrt z$ has a branch point at $\infty$ and is bounded in a neighborhood of $\infty$.