0
$\begingroup$

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

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

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 $z \neq 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.