User:Sam Derbyshire/Test

$f(z)={\frac {e^{z}}{z^{5}-z^{4}+2z^{3}-2z^{2}+z-1}}$

$z^{5}-z^{4}+2z^{3}-2z^{2}+z-1=(z^{2}+1)^{2}(z-1)=(z-i)^{2}(z+i)^{2}(z-1)$

And $e^{z}\not =0$

As such, $f(z)={\frac {e^{z}}{(z-i)^{2}(z+i)^{2}(z-1)}}$ has a pole of order 1 at z = 1 and two poles of order two at z = i and z = -i.

Let $\gamma$ be some Jordan curve around all these three poles : for example, the curve parametrized by $2e^{it}$ with $0\leq t\leq 2\pi$.

Then $\oint _{\gamma }f(z)dz=2\pi i{\Big (}\operatorname {Res} _{z=i}f(z)+\operatorname {Res} _{z=-i}f(z)+\operatorname {Res} _{z=1}f(z){\Big )}$ by the residue theorem.

$\operatorname {Res} _{z=1}(f(z))={\frac {1}{2\pi i}}\oint _{\gamma _{1}}f(z)dz={\frac {1}{2\pi i}}\oint _{\gamma _{1}}{\frac {e^{z}}{(z-i)^{2}(z+i)^{2}(z-1)}}={\frac {1}{2\pi i}}\oint _{\gamma _{1}}{\frac {\frac {e^{z}}{(z-i)^{2}(z+i)^{2}}}{z-1}}dz$

But the residue of a function f(z) at a simple pole c is given by $\operatorname {Res} _{z=c}f(z)=\lim _{z\rightarrow c}(z-c)f(z)$.

Thus ${\frac {1}{2\pi i}}\oint _{\gamma _{1}}{\frac {\frac {e^{z}}{(z-i)^{2}(z+i)^{2}}}{z-1}}dz=\lim _{z\rightarrow 1}{\frac {e^{z}}{(z-i)^{2}(z+i)^{2}}}={\frac {e^{1}}{(1-i)^{2}(1+i)^{2}}}={\frac {e^{1}}{(-2i)(2i)}}={\frac {e}{4}}$.

Then $\operatorname {Res} _{z=1}f(z)={\frac {e}{4}}$

$\operatorname {Res} _{z=i}(f(z))={\frac {1}{2\pi i}}\oint _{\gamma _{i}}{\frac {\frac {e^{z}}{(z+i)^{2}(z-1)}}{(z-i)^{2}}}dz$.

This time, the pole is of second order, thus its residue is given by the formula :

$\operatorname {Res} _{z=c}f(z)={\frac {1}{(n-1)!}}\lim _{z\rightarrow c}\left({\frac {d}{dz}}\right)^{n-1}\left(f(z)\cdot (z-c)^{n}\right)$, where n is the order of the pole.

Thus, $\operatorname {Res} _{z=i}f(z)=\lim _{z\rightarrow i}{\frac {d}{dz}}{\big (}f(z)(z-i)^{2}{\big )}=\lim _{z\rightarrow i}{\frac {d}{dz}}{\bigg (}{\frac {e^{z}}{(z+i)^{2}(z-1)}}{\bigg )}=\lim _{z\rightarrow i}{\frac {e^{z}(z^{2}-4z+iz+2-2i)}{(z+i)^{3}(z-1)^{2}}}={\frac {3ie^{i}}{8}}$.

$\operatorname {Res} _{z=-i}(f(z))={\frac {1}{2\pi i}}\oint _{\gamma _{-i}}{\frac {\frac {e^{z}}{(z-i)^{2}(z-1)}}{(z+i)^{2}}}dz$.

Following the same procedure :

$\operatorname {Res} _{z=-i}f(z)=\lim _{z\rightarrow -i}{\frac {d}{dz}}{\big (}f(z)(z+i)^{2}{\big )}=\lim _{z\rightarrow -i}{\frac {d}{dz}}{\bigg (}{\frac {e^{z}}{(z-i)^{2}(z+1)}}{\bigg )}=\lim _{z\rightarrow -i}{\frac {e^{z}(z^{2}-4z-4iz+2+2i)}{(z-i)^{3}(z+1)^{2}}}=-{\frac {3ie^{i}}{8}}$.

So <math>\oint_{\gamma} f(z) dz = 2 \pi i \Bigg( \frac{e}{4} + \frac{3ie^i}{8} - \frac{3ie^i}{8} \Bigg) = \frac{e \pi i}{2}.