Calculate integral $$\oint\limits_{\gamma}\frac{e^z}{z^4+5z^3}dz$$ Where $\gamma$ is parameterization of one rotation of circle $A(0,2)$
So if I write the integral like this $$\oint\limits_{\gamma}\frac{e^z}{z^3(z-5)}dz$$ We see that there is simple pole at $z=(5,0)$ and pole order of $3$ at $(0,0)$.
So isn't residue for the simple pole. Out side of $\gamma$. \begin{align}\operatorname{Res}\limits_{z_0=5}&= \lim_{z\to 5} (z-5) f(z)\\ &= \lim_{z\to 5}(z-5) \frac{e^z}{z^3 (z-5)} \\ &=\lim_{z\to 5} \frac{e^z}{z^3}\\ &=\frac{e^5}{125} \end{align} And for the origo $(k=3)$ \begin{align} \operatorname{Res}\limits_{z_0=0}&= \frac{1}{(k-1)!}\lim_{z\to 0} \frac{\partial^{k-1}}{\partial z^{k-1}}(z-0)^k f(z)\\ &= \frac 12 \lim_{z\to 0} \frac{\partial^{2}}{\partial z^2}z^3\frac{e^z}{z^3 (z-5)} \\ &=\frac 12 \lim_{z\to 0} \frac{e^z ( (z-5)^2 - 2(z-6))}{(z-5)^2}\\ &=-\frac{37}{250} \end{align}
So the integral value should be $$2\pi i (- 37/250) =-\frac{74 \pi i}{250}$$
Is this the right way to find the integral or should I have found the laurent series for $f(z)$ or something
It is simplier with the Laurent Formula indeed:
the only residue inside $\gamma$ is for $z=0$.
$$ \frac{\exp z}{(z-5)z^3} = -\frac 15\sum_{m=0}^\infty \frac{z^m}{m!} \sum_{n=0}^\infty z^{n-3}5^{-n} $$ hence the residual is $$ -\frac 15 \left( \frac1{0!} 5^{-2} + \frac1{1!} 5^{-1} + \frac1{2!} 5^{-0} \right) = -\frac 15\times\frac{37}{50} $$
and the integral is $$\oint\limits_{\gamma}\frac{e^z}{z^4+5z^3}dz = -\frac{37\pi}{125}$$