I realize this is not the fastest way of getting a Taylor's series expansion of $f(z)=\log(z)$ about $z=1$. But here goes. I am assuming I am working on the principal branch of the logarithm ($-\pi<\theta<\pi$). I am assuming that $f(1)=\log(1)=0$. That's the branch I am on.
Next, some derivatives:
And checking, $f(1)=0$, which is my assumption above.
Question: Now, suppose I start the problem again, only this time I assume I am a different branch of the logarithm. Let's assume that this time we are working on the branch $\pi<\theta<3\pi$, so that $f(1)=\log(1)=2\pi i$.
If I repeat the calculations above, what would they now look like, and what would be the answer for my series?