<s>[INST] Power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ I want to find the power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ and to find the radius of convergence, what does this take?
Is this just a taylor series with $z=4i$ subbed in? What about finding radius of convergence?
 [/INST] This is an expansion in powers of $(z-4i)$, so the first step is to locate $z-4i$ in this expression.
$$
\frac 1 {3-((z-4i)+4i)} = \frac 1 {(3-4i) - (z-4i)}.
$$
To get it into the form $\dfrac 1 {1-r}$ we need to divide the numerator and denominator by the constant term in the denominator, which is $3-4i$:
\begin{align}
& \frac 1 {(3-4i) - (z-4i)} = \frac {\frac 1 {3-4i}} {\frac{3-4i}{3-4i} - \frac{z-4i}{3-4i}} = \frac {\frac 1 {3-4i}} {1 - \frac{z-4i}{3-4i}} = \frac {3+4i} {25} \cdot\frac 1 {1-\frac{z-4i}{3-4i}} \\[10pt]
= {} & \frac{3+4i} {25} \cdot \frac 1 {1-r} = \frac{3+4i} {25} \left( 1+r + r^2 + r^3 + \cdots \right) \\[10pt]
= {} & \frac{3+4i} {25} \left( 1 + \frac1{3-4i}(z-4i) + \frac1{(3-4i)^2}(z-4i)^2 + \frac1{(3-4i)^3}(z-4i)^3 + \cdots \right) \\[10pt]
= {} & \frac{3+4i} {25} \left( 1 + \frac{3+4i} {25} (z-4i) + \left(\frac{3+4i} {25} \right)^2 (z-4i)^2 + \left(\frac{3+4i} {25} \right)^3 (z-4i)^3 + \cdots  \right)
\end{align}
This converges when $|r|<1$, i.e. when $\left|\dfrac{z-4i}{3-4i}\right|<1$, i.e. when $|z-4i|<|3-4i|=5$.  Thus the radius of convergence is $5$, and the center of the circle of convergence is $4i$.  The number $3$ is exactly on the boundary of the disk of convergence.  So are $-i$, $-3$, and $9i$.
</s>