Finding $\oint_{|z|=a}\frac{e^{3z}}{z^{4}-6z^{3}+13z^{2}}dz$

Question

$$\oint_{|z|=a}\frac{e^{3z}}{z^{4}-6z^{3}+13z^{2}}dz$$

I tried to solve $z^{4}-6z^{3}+13z^{2}=0$ and I got the poles $0,3+2i,3-2i$ How to continue?

Answer 1

Residue theorem states that if $D \subset \mathbb{C}$ open with $f : D - S \longmapsto \mathbb{C}$ holomorphic, $S$ closed and discrete in $D$. Let $R \subset D$ compact e with piecewise $C^{1}$ boundary, so $R \cap S = \left\lbrace z_{1},\cdots, z_{k} \right\rbrace$ is finite. $\partial R \cap S = \emptyset$ then $\int_{\partial R}f(z)dz = 2\pi i \sum\limits_{1 \leq i \leq k} Res(f,z_{i})$

To evalute simple poles remember that the residue in a point is the first term on the Laurent expansion of $f$, i.e $a_{-1}$, in other words $f(z) = \frac{a_{-1}}{z-z_{0}} + \sum\limits_{n \geq 0 } a_{n}(z-z_{0})^{n}$.

So $(z-z_{0})f(z) = a_{-1} + \sum\limits_{n \geq 0 } a_{n}(z-z_{0})^{n}$ taking the limit for $z \to z_{0}$ we have $Res(f,z_{0}) = a_{-1} = \lim\limits_{z \to z_{0}}f(z)(z-z_{0})$. From here you can notice that if $f = \frac{p}{q}$ as in this case sobstituing in the limit we got $a_{-1} = \frac{p(z_{0})}{q'(z_{0})}$, well define since $z_{0}$ is a simple pole.

So the integral is easily computable, do you see what $R$ should be in order to correctly apply the Residue theorem ?

Answer 2

Before we get started, note the integrand would only have real-valued coefficients if we expand it as a Laurent series, so the integral must be $2\pi i$ times something real, i.e. imaginary. (In fact, the $z=0$ pole's contribution to the integral will be $\pi i$ times a rational number.) Unfortunately, this doesn't give us much of a sanity check, so you'll want to very carefully check all my arithmetic.

The pole at $0$ is second-order, so its residue is$$\lim_{z\to0}\frac{d}{dz}\left(\frac{e^{3z}}{z^2-6z+13}\right)=\lim_{z\to0}\left(\frac{3e^{3z}}{z^2-6z+13}-\frac{(2z-6)e^{3z}}{(z^2-6z+13)^2}\right)=\frac{45}{169}.$$The pole at $3\pm2i$ is first-order, so its residue is$$\lim_{z\to3\pm2i}\frac{e^{3z}}{z^2(z-3\pm2i)}=\frac{(\tfrac54\sin6-3\cos6\mp i(3\sin6+\tfrac54\cos6))e^9}{169}.$$The integral's Cauchy principal value is $2\pi i$ times the sum of the residues of $|z|<a$ poles, plus $\pi i$ times the sum of the residues of $|z|=a$ poles. Since $a>0$, the $z=0$ pole always contributes $\frac{90\pi i}{169}$. If $a>\sqrt{13}$, we add the quantity below to this (but only half as much if $a=\sqrt{13}$, the only choice of $a$ for which we need to use a Cauchy PV):$$2\pi i\sum_\pm\frac{(\tfrac54\sin6-3\cos6\mp i(3\sin6+\tfrac54\cos6))e^9}{169}=\frac{\pi i(5\sin6-12\cos6)e^9}{169}.$$So regardless of $a$, the integral is imaginary-valued, as expected.