The Problem: Find the Taylor series for the logarithm branch $0<\arg(z)<2 π$ in powers of $z+2$
My resolution:
The method I've used to come up with an answer was integration of a known series. I know that the series of $f(z)= 1/z$ around the point $z=-2$ is $\sum_{n=0}^{∞} \frac{-(z+2)^n}{2^{n+1}}$ which converges if $|z+2|<2$
So by doing this I have:
$\frac{1}{z}$ = $\sum_{n=0}^{∞} \frac{-(z+2)^n}{2^{n+1}}$
Since I know that $1/z$ is an antiderivative of $\log(z)$ when $0<\arg(z)<2\pi$ (a branch is specified), I can integrate both sides of the equation to obtain:
$\log_0(z)= \sum_{n=0}^{∞} \frac{-(z+2)^{n+1}}{(n+1)(2^{n+1})}$
My main questions are:
-Is this approach a correct one?
-What would change If I had a different branch of $log(z)$ ?
Thank you.
Answering your second question:
For any branch of $\log z$, $$(\log z)’=\frac1z$$ which is single-valued.
Thus, if $$\log z=\sum^\infty_{n=0}a_n (z-c)^n$$ we expect $(\log z)’=\sum^\infty_{n=1}a_n\cdot n (z-c)^{n-1}$ should be independent of choice of branch.
In other words, $a_1,a_2,...$ remain the same even if you change your choice of branch.
The choice of branch can only affects $a_0$, which can be uniquely determined by evaluating $\log c=\ln|c|+i\arg c$, with the $\arg c$ that fits with the choice of branch.
For the first question:
Substitute in $z=-2$ into your series, and you will obtain $\log_0(-2)=0$, which is absurd.
You made a mistake: missing the integration constant.