I am trying to prove $f(z)=\sqrt{2z-2\log(z)-2}$ is analytic near $z=1$. The issue is proving there is no branch point.
If I try the approach of taking the path $z=1+r\exp(i\theta)$ with $r=\epsilon$ and $\theta$ varying from $0$ to $2\pi$, I'm finding it hard to show that the value did not change:
For $f(r,\theta)$, $f(\epsilon,0) = \sqrt{2r\exp(i0)-2\log(1+r\exp(i0))} = \sqrt{2r-2\log(1+r)}$ and $f(\epsilon,2\pi) = \sqrt{2r\exp(i2\pi)-2\log(1+r\exp(i2\pi))} = \,\,??$.
Not sure if this is the right approach, but it's how I learned to do it. Any advice?