The Taylor series of $f(z) := \log z$ about $z_0 = -1 + i$ So the problem states:

Say $f(z) := \log z$ is the principal branch of the logarithm (the primitive of $1/z$ on the region $\Bbb C\setminus (-\infty,0]$). Show that the Taylor series of $f(z)$ about $z_0 = -1 + i$ takes the form
  $$\log z = \sum_{n=0}^{\infty} a_n(z-(-1+i))^n $$
  with
$$a_0 = \log \sqrt{2} + i\frac{3\pi}{4}\,\,\,\text{and}\,\,\,a_n = (-1)^{n+1}\frac{e^{-3\pi in/4}}{n2^n/2}$$
Determine the radius of convergence of this series. Explain why the series does not represent $f(z)$ in its entire disk of convergence.

My main concern here is how to show $\log(-1+i) = \log \sqrt{2} + i\frac{3\pi}{4} $ and determine the radius of convergence.
 A: The function $g(z):={1\over z}$ is analytic in $\dot{\mathbb C}:={\mathbb C}\setminus\{0\}$, but  has no primitive defined in all of $\dot{\mathbb C}$.
The function $g$ however has primitives in suitable subdomains $\Omega\subset\dot{\mathbb C}$, the most famous one being the principal value
$${\rm Log}(z):=\log|z|+ i\>{\rm Arg}(z)\ ,$$
which is defined on $\Omega:={\mathbb C}\setminus\{≤0\ {\rm real\ axis}\}$. In particular
$${\rm Log}(-1+i)={1\over2}\log2+{3\pi\over4}\>i\ .$$
Standing at the point $p:=-1+i\in\dot{\mathbb C}$ we see that the function $g$ is analytic in a disk $D$ of radius $\sqrt{2}$ around $p$. Therefore $g$ has primitives which are analytic in $D$, whence have a power series development with center $p$ and convergence radius $\sqrt{2}$. These primitives are equal up to an additive  constant, and one of them has the value ${1\over2}\log2+{3\pi\over4}\>i$ at $p$. Since ${\rm Log}$ is a primitive of $g$ in the neighborhood of $p$ having exactly this value at $p$ the corresponding series represents ${\rm Log}$ in any domain $\Omega'\subset D\cap\Omega$ containing the point $p$. The points of $\Omega$ lying below the negative real axis do not belong to such an $\Omega'$. Therefore the obtained series does not represent ${\rm Log}$ there.
A: The first part of your question is really asking how to find the real and imaginary parts of the logarithm.  So just write it out!
$$
e^{x+ i y} = -1 + i
$$
Now use Euler's identity to get simultaneous equations for $x$ and $y$:
$$
e^x \cos y = -1 ~,~ e^x \sin y = 1
$$
Hopefully you can solve these (the solution for $y$ is only unique because we choose a particular branch of the logarithm).
As for the second part, what do you know about the radius of convergence of the Taylor series for a holomorphic function?
A: The Maple command $$convert(ln(z), FPS, z = -1+I)  $$ produces $$ \ln \left( -1+i \right) +\sum _{k=0}^{\infty } \left( -\frac {
 \left( 1/2+1/2\,i \right)^k}{2\,k+2}-\frac {i \left( 1/2+1/2\,i
 \right)^k}{2\,k+2} \right)  \left( z+1-i \right) ^{k+1}
 .$$ Next,
$$normal(-(1/2+1/2*I)^k/(2*k+2)-I*(1/2+1/2*I)^k/(2*k+2)) $$ outputs $$\frac{\left( -1/2-1/2\, i \right) \left( 1/2+1/2\,i \right)^k} {k+1}.
 $$
