Determine a branch of $f(z)=\log(2iz-z^2)$ that is analytic at $z=1$. Then find $f(1)$ and $f'(1)$.
First we note that $g(z)=2iz-z^2$ and recall $\mathcal{L}_\tau:=\log|z|+i\arg_\tau z$. So a branch we want would be $F(z)=\mathcal{L}_\tau(g(z))$. However, I'm having a hard time understand how to actually do this. I think I have to start by saying $$g(z)=0\implies 2iz-z^2=0\implies z=0, z=2i$$. OR I have to insert $1$ since that's what we want, which is to say $$g(1)=2i-1$$ But then I'm unsure of what to do with this information or where to go from here. The $\tau$ and $\mathcal{L}$ really makes stuff wierd, especially $i\arg_\tau z$, because $\arg z$ is something different.
When we write out $f(1)$ for the single-valued function with branch, it's
$$ f(1) = \mathcal{L}_\tau(g(1)) = \log |2i-1| + i \arg_\tau(2i-1) $$
So $\tau$ just needs to be chosen so that $g(1) = 2i-1$ is not on the branch cut. The usual branch cut along the negative reals with $-\pi < \arg z \leq \pi$ will work fine since $2i-1$ is not real.