# Finding power series and ROC of complex function

I have the function $f(z) = \frac{3iz-6i}{z-3}$

I need to find a power series $\sum c_n (z-1)^n$ about $z_0 = 1$

I can rewrite $f$ as $\frac{2i-iz}{1-\frac{z}{3}}$, where I'm guessing the ROC would then be $3$.

However I have it in the form $\frac{a}{1-r}$ and I'm not sure with $a$ being dependent on $z$ that I can get a correct power series, also I don't see how we get his power series about $1$.

I've considered Taylor Series, but it seems far too complicated with complex numbers.

I've tried writing it as $3i \frac{z-2}{z-3}$ where now I just need to get the power series for $\frac{z-2}{z-3}$ but still having trouble.

Any ideas what a general approach might be. I'm more interested in the method over the answer.

EDIT:

I've also tried writing it as

$3i(\frac{1}{1-\frac{3}{z}})+2i (\frac{1}{1-\frac{z}{3}})$

which would get me ROC = $3$ again. But I'm not sure how this would be about $z_0 = 1$

• You want a power series with centre $1$. Writing $z = w+1$ may help finding it. May 10, 2016 at 19:29
• Can you explain a little further. Sorry if I'm a bit confused May 10, 2016 at 19:44
• $w$ is $z-1$. So you want an expansion in terms of $w$. If you had for example $\frac{1}{z-5}$ and wanted to expand it as a power series in $z-2$, you'd set $z = w+2$, and in terms of $w$ the function becomes $\frac{1}{(w+2)-5} = \frac{1}{w-3} = -\frac{1}{3}\cdot \frac{1}{1-w/3}$. You know how to expand that into a geometric series (with common ratio $w/3$), and then you just replace $w$ with $z-2$. May 10, 2016 at 19:52
• Thanks Daniel. From this I get $3i(\frac{1}{1-\frac{2}{w}})-\frac{3i}{2}(\frac{1}{1-\frac{w}{2}})+3i(\frac{1}{1-\frac{w}{2}})$. However I'm not sure how to get the ROC for $\frac{1}{1-\frac{2}{w}}$ May 11, 2016 at 9:11

The function $f(z)=\frac{3iz-6i}{z-3}$ has a simple pole at $z=3$. We want to develop $f$ into a power series at $z_0=1$. Since the ROC is the distance from the center of the series to the nearest singularity or branch point we obtain $ROC=2$.
The idea is to transform the function to use a geometric series representation with center $z=z_0$. \begin{align*} \frac{1}{z-a}&=-\frac{1}{(a-z_0)-(z-z_0)}\\ &=-\frac{1}{a-z_0}\cdot\frac{1}{1-\frac{z-z_0}{a-z_0}}\\ &=-\frac{1}{a-z_0}\sum_{n=0}^{\infty}\frac{1}{(a-z_0)^n}(z-z_0)^n\qquad\qquad|z-z_0|<|a-z_0| \end{align*}
We obtain for $|z-1|<2$ \begin{align*} f(z)&=\frac{3iz-6i}{z-3}\\ &=-\frac{3iz-6i}{2-(z-1)}\\ &=-\frac{3iz-6i}{2}\cdot\frac{1}{1-\frac{z-1}{2}}\\ &=3i\frac{1-(z-1)}{2}\sum_{n=0}^{\infty}\frac{1}{2^n}(z-1)^n\\ &=\cdots \end{align*}