Integrate $\frac{z^3 . e^\frac{1}{z}}{z+1}$ around $\vert z \vert = 2$

Question

My attempt:

Let $f(z)= \frac{z^3 . e^\frac{1}{z}}{z+1}$. There are two singularities inside the curve, $z_0=0$ and $z_1=-1$ so by using Cauchy's theorem I'm going to split the integral and integrate around both curves.

Around $z=-1$, I take the curve $\gamma_1 : -1 + \frac{1}{2}e^{it}$ where $t \in [0, 2\pi]$. Note that $-1 \in $ int $\gamma_1$. Therefore, by letting $g(z)=z^3 . e^\frac{1}{z}$, we see that $g(z)$ is analytic in the region bounded by $\gamma_1$ and thus I can claim that $\int_{\gamma_1} f(z) dz = g(-1).2\pi.i = -\frac{2.\pi.i}{e}$.

Around $z=0$, I use the residue formula that states that $Res_{z=0} = \frac{1}{z^2} f(\frac{1}{z}) = \frac{1}{2.\pi.i} \int_{\gamma_2} f(z)dz$, taking $\gamma_2 : \frac{1}{2}e^{it}$ where $t \in [0, 2\pi]$ to ensure $f(z)$ is analytic in the region bounded by $\gamma_2$ except at $z=0$. Next, I rewrite $\frac{1}{z^2} f(\frac{1}{z}) = \frac{e^z}{z^4}.\frac{1}{1-(-z)}$.

Note that I'm interested in the coefficient of $z^{-1}$ so I expand both series.

$\frac{e^z}{z^4}.\frac{1}{1-(-z)} = \sum_{k=0}^\infty \frac{z^k}{k!.z^4} . \sum_{n=0}^\infty (-z)^n$. Note that the series with index $k$ has powers of $z$ beginning with -4, and the series with index $n$ has powers of $z$ beginning with 0. So, I extract all the terms that will multiply to $z^{-1}$.

$[\frac{1}{0!z^4} + \frac{1}{1!z^3} + \frac{1}{2!z^2} + \frac{1}{3!z} + ...][... + (-1)^3z^3+(-1)^2z^2+(-1)^1z^1+(-1)^0z^0]$

Multiplying relevant terms, I get $\frac{-2}{3z}$ therefore the residue is $\frac{-2}{3}$ and the integral is $-\frac{4\pi.i}{3}$.

Adding up both integrals I get $\frac{-2\pi.i}{e}+\frac{4\pi.i}{3}$.

My questions are:

1. Is this a correct attempt?
2. Is there any shorter way to do this integral?

Answer 1

We can use the full residue theorum: $$\oint_{\gamma}f(z)dz=2{\pi}i\sum{Res(f,a_k)}$$ Therefore doing both those integrals in one.

For "simple poles" such as $z=-1$ in this example, we can actually calculate residues using a really simple limit: $$Res(f,c)=\lim_{z{\rightarrow}c}(z-c)f(z)$$ Where $c$ is our simple pole. Of course this is just an idea formed from Cauchy's original theories. If you need further information on types of singularities and how we compute residues for each type this is a good link.

Therefore: $$Res(f,-1) = \lim_{z{\rightarrow}-1}(z-(-1))\frac{z^3e^{\frac{1}{z}}}{z+1}$$ $$=\lim_{z{\rightarrow}-1}z^3e^{\frac{1}{z}}=-\frac{1}{e}$$

So now we just need our other residue, which unfortunately is from an essential singularity. As far as I am aware, your method using the Laurent series expansion is the best way to compute that.

Therefore, the solution (using your residue for $z=0$ which I hope is right!) is: $$\oint_{\gamma}f(z)dz=2{\pi}i(-\frac{1}{e}-\frac{2}{3})=\frac{-2{\pi}i}{e}-\frac{4{\pi}i}{3}$$ As you got!

The residue theorum is all derived from Cauchy's theorum, so you can see the obvious similarities, but lets us combine the contour integrals into $2{\pi}i$ multiplied by the sum of all the residues inside the closed contour. Just keep in mind what that closed contour is and which singularities it encloses.