# Find the Laurent series expansion in powers of z

Find the Laurent series expansion in powers of $z$ of

$$f(z)=\frac{e^{2z}} {z}$$

valid in the region $|z|>$0.

Any help appriciated. Thanks

• Did you mean $\lvert z\rvert > 1$? The function has only one singularity in $\mathbb{C}$, at $1$. – Daniel Fischer May 19 '14 at 14:15
• Sorry i mean z, not z-1 so The singularity is at 0 – user134400 May 19 '14 at 14:47

The Maclaurin series of $e^{2z}$ is
$$1 + (2z) + \frac{(2z)^2}{2!} + \cdots = \sum_{k=0}^\infty \frac{2^k z^k}{k!}$$ so the Laurent series you're looking for is simply $1/z$ times this, i.e.: $$\sum_{k=0}^\infty \frac{2^k z^{k-1}}{k!} = \sum_{k=-1}^\infty \frac{2^{k+1} z^{k}}{(k+1)!}.$$
The Maclaurin series of $e^{2z}$ converges on all of $\mathbb{C}$, since $e^{2z}$ is entire. Hence the Laurent series above converges for $z \neq 0$.