I've done a little math and I got the following power series expansion of $\log z$ about $z_0=-2+i$.

$$\log z=\log(-2+i)+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(-2+i)^n}[z-(-2+i)]^n$$

I've shown that the series converges on $|z-(-2+i)|<\sqrt5$ but I think it only agrees with $\log z$ on $|z-(-2+i)|<1$.

I have:

f = Function[z, 
  Log[-2 + I] + 
   Sum[(-1)^(n + 1)*(z - (-2 + I))^n/(n*(-2 + I)^n), {n, 1, 
     Infinity}]]

And

f[-2 + I/2]

produces

log(-2+I)+log(9/10+I/5)

I'm wondering if there is an easy way to force this as a logarithm of a single number. I'm also wondering if someone has a suggestion on how to explore the convergence on the two regions defined above.

I've done a little math and I got the following power series expansion of $\log z$ about $z_0=-2+i$.

$$\log z=\log(-2+i)+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(-2+i)^n}[z-(-2+i)]^n$$

I've shown that the series converges on $|z-(-2+i)|<\sqrt5$ but I think it only agrees with $\log z$ on $|z-(-2+i)|<1$.

I have:

f = Function[z, Log[-2 + I] + 
      Sum[(-1)^(n + 1)*(z - (-2 + I))^n/(n*(-2 + I)^n), {n, 1, Infinity}]]

The evaluation

f[-2 + I/2]

produces

log(-2 + I) + log(9/10 + I/5)

I'm wondering if there is an easy way to force this to a logarithm of a single number. I'm also wondering if someone has a suggestion on how to explore the convergence on the two regions defined above.