Locate the poles and the find the residues of the following function:

$\dfrac {e^{5z}}{\left( z+i\pi \right) ^{3}}$

From my understand there should be a pole at $-i\pi$ so y the residue theorem we have :

$Res\left\{ f,\alpha \right\} = \dfrac {g\left( \alpha \right) }{h'\left( \alpha \right) }$

But since theres a singularity at $-i\pi$ surely the residue would just be dividing by zero as $h'\left( \alpha\right)$ is zero ?

  • $\begingroup$ Be careful it is not a simple pole. $\endgroup$ – Jennifer May 7 '16 at 13:57
  • $\begingroup$ Can you expand on that please $\endgroup$ – Tom Otto May 7 '16 at 13:58
  • $\begingroup$ $Res\left\{ f,\alpha \right\} = \dfrac {g\left( \alpha \right) }{h'\left( \alpha \right) }$ is true when $h$ has a zero of order $1$ but here it is of order $3$, you must have an other formula more general in your book. $\endgroup$ – Jennifer May 7 '16 at 14:02

For a pole $c$ of order $n$, the general formula is : $$\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right)$$

Here we have : $$\begin{align} \mathrm{Res}(f,-\pi i)&=\frac{1}{(3-1)!} \lim_{z \to - \pi i} \frac{d^{3-1}}{dz^{3-1}}\left( (z+\pi i)^{3}\dfrac {e^{5z}}{\left( z+i\pi \right) ^{3}} \right)\\&=\frac{1}{2} \lim_{z \to - \pi i} \frac{d^{2}}{dz^{2}}\left( e^{5z} \right)\\&=\frac{1}{2} \lim_{z \to - \pi i} 25e^{5z}\\&= \frac{25}{2}e^{-5\pi i}\end{align}$$

Finally : $\mathrm{Res}(f,-\pi i)=-\frac{25}{2}$


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