# Taylor expansion of a complex function on a disc

I need to find the taylor expansion of the complex function $\frac{z^2}{z-2}$ on the disc $|z|<2$

I'm not sure how to start this off, can anyone help me?

• A Hint: Do you know the formula for the geometric series? – mlk Dec 1 '15 at 21:54

Hint: You can find the taylor expansion of $\frac{1}{z-2}$ using geometric series. With some algebraic manipulation, you can use this to find the taylor expansion of $\frac{z^2}{z-2}$.
Looks to me like straight forward division. z- 2 divides into z^2+ 0z+ 0 z times with remainder 2z: $\frac{z^2}{z- 2}= z+ \frac{2z}{z- 2}$. Then z- 2 divides into 2z twice with remainder 4: $\frac{2z}{z- 2}= 2+ \frac{4}{z- 2}$ so $\frac{z^2}{z- 2}= z+ 2+ \frac{4}{z- 2}= (z- 2)+ 4+ \frac{4}{z- 2}$
• Take a look at the other answers -- you now need the T.S of $4/(z-2)$... – peter a g Dec 1 '15 at 22:06
$$f(z)=\frac{1}{z-2}=-\frac{1}{2}-\frac{z}{4}-\frac{z^2}{8}-...$$ $$\implies z^2f(z)=\frac{z^2}{z-2}=-\frac{z^2}{2}-\frac{z^3}{4}-\frac{z^4}{8}-...$$