Find the residue of $\frac{e^{iht}}{(t-i)^2}$

Question

I am trying to find the residue of

$\frac{e^{iht}}{(t-i)^2}$

At $t = i$, $h$ is a constant.

First of all, how do I detect which way is faster? Using the residue formula or expanding the function into Laurent series?

Then how would you do it?

Thanks

Answer 1

$f(t)=e^{iht}$ is analytic everywhere then the Laurent series at $z=i$ is the Taylor series and $f^{(n)}(i)=(ih)^ne^{-h}$

Then, if $|t-i| < \infty$

$$ e^{iht}= \sum_{n=0}^ \infty \frac{(ih)^ne^{-h} (t-i)^n}{n!} = \frac{1}{e^h}+\frac{ih (t-i)}{e^h} +\ldots+ $$

Then, if $0<|t-i|<\infty$

$\displaystyle \frac{e^{iht}}{(t-i)^2}= \frac{1}{e^h(t-i)^2}+\frac{ih}{e^h(t-i)}+ \ldots+$

Then the residue is $ \displaystyle \frac{ih}{e^h}$

Answer 2

In general, $$ \operatorname{Res}(f;a)=\frac{1}{(n-1)!}\lim_{z\to a} \frac{d^{n-1}}{dz^{n-1}}((x-a)^nf(z)), $$ where $n$ is the order of the pole $a$. Thus, $$ \operatorname{Res}\left(\frac{e^{iht}}{(t-i)^2};i\right)=\lim_{t\to i} \frac{d}{dz}e^{iht}=ihe^{iht}\mid_{t=i}=ihe^{-h}. $$