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?
 A: 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 $9i$.
A: i will use the fact that a geometric series with first term $a$ and common ration $r$ is $$\frac{a}{1-r} = a + ar + ar^2 + \cdots, \text{ for }|r| < 1 $$
we will make a change of variable $h = z- 4i, z = h + 4i$, then $$\frac{1}{3-z} = \frac1{3-4i - h} = \frac{1/(3-4i)}{1-h/(3-4i)} = \frac{1}{3-4i}+\frac{h}{(3-4i)^2 } + \cdots +|h|/|3-4i|=|h|/5 < 1 $$
the radius of convergence is $5.$ and $$\frac{1}{3-z} = \frac{1}{3-4i}+\frac{(z-4i)}{3-4i^2 } + \cdots \text { for }  |z - 4i|<5. $$
A: Remember the geometric series: $\sum \limits _{n=0} ^\infty z^n = \frac 1 {1-z}$ for $|z|<1$. Let us use this to solve your problem.
If you want to write the Taylor series around $4 \Bbb i$, you need to "fabricate" a power series in $z-4\Bbb i$, so let us produce it. One approach would be to compute the generic $n$-th derivative of $f$ and then write down the usual Taylor series. I'll leave this to you, it is not difficult ($f^{(n)} (z)= \frac {(-1)^n n!} {(z-4 \Bbb i)^{n+1}}$).
The alternative approach is as follows: $$f(z) = \frac 1 {3-z} = \frac 1 {3 - (z - 4 \Bbb i) - 4 \Bbb i} = \frac 1 {3 - 4 \Bbb i} \frac 1 {1 - \frac {z-4 \Bbb i} {3 - 4 \Bbb i}} = \frac 1 {3 - 4 \Bbb i} \sum \limits _{n=0} ^\infty \Big( \frac {z-4 \Bbb i} {3 - 4 \Bbb i} \Big)^n = \sum \limits _{n=0} ^\infty \frac 1 {(3 - 4 \Bbb i) ^{n+1}} (z-4 \Bbb i)^n ,$$
where the sum has appeared thanks to the formula of the gometric series with $z$ replaced by $\frac {z-4 \Bbb i} {3 - 4 \Bbb i}$, provided that $|\frac {z-4 \Bbb i} {3 - 4 \Bbb i}| <1$, i.e. $|z-4 \Bbb i| < 5$. This last result also tells you that the radius of convergence is $5$.
