Laurent series of a function

Question

I have problems showing the Laurent series of: $f(z)$$=$$1/(z+a)$ For any $a$ in $C$ centred at $z=0$.

Answer 1

We assume $a\neq0$.

One may write, for $|z|<|a|$, using a standard geometric series expansion, $$ \frac1{z+a}=\frac1a \cdot \frac1{1+\frac{z}a}=\frac1a \cdot \sum_{n=0}^\infty (-1)^n\frac{z^n}{a^n}. $$

Answer 2

You may use the geometric series, since $|z| \to 0$, which is your point.

$$\frac{1}{z+a} = \frac{1}{a(\frac{z}{a} + 1)} = \frac{1}{a}\frac{1}{1 + \frac{z}{a}} = \frac{1}{a}\sum_{k = 0}^{+\infty}\left(-\frac{z}{a}\right)^k$$

That is

$$\sum_{k = 0}^{+\infty}(-1)^k \frac{z^k}{a^{k+1}}$$

Clearly this holds for $|z| < |a|$, and $a\neq 0$.