Use Cauchy's theorem to prove that $\int_{C}{\frac{z^2+2z-5}{(z^2+4)(z^{2}+2z+2)}dz}=0$

Question

Let $C$ the circle $|z|=r$, show that $$\lim_{r\to\infty}\int_{C}{\frac{z^2+2z-5}{(z^2+4)(z^{2}+2z+2)}dz}=0$$

My approach: Note that $|z^2+2z-5|\leq r^2+2r+5$ and $|(z^2+4)(z^{2}+2z+2)|=|(z^2+4)||(z^{2}+2z+2)|\leq (r^2+4)(r^2+2r+2)$. So, $$\left|\frac{z^2+2z-5}{(z^2+4)(z^{2}+2z+2)}\right|\leq \frac{r^2+2r+5}{(r^2+4)(r^2+2r+2)}\to 0$$

When $r\to\infty$.

However, how apply Cauchy's theorem in this case; This theorem is usually formulated for closed paths as follows, let $U$ be an open subset of $\mathbb{C}$ which is simply connected, let $f : U → \mathbb{C}$ be a holomorphic function, and let $\gamma$ be a rectifiable path in $U$ whose start point is equal to its end point. Then $$\int_{\gamma}{f(z)dz}=0$$ I know that, If $f(z)$ is analytic in a bounded region, limited by closed simple curve $C_1,C_2,\ldots, C_n$ and on this curves, then $$\int_{C}{f(z)dz}=\int_{C_{1}}{f(z)dz}+\int_{C_{2}}{f(z)dz}+\ldots+\int_{C_{n}}{f(z)dz}$$ And in this case the singularities points are $\{-1-i,-1+i,-2i,2i\}$, so we can construc curves around this points, can give me any hint, thanks!

Answer 1

The first argument can be easily made to work, we have for $r>10$ the estimations: $$ \begin{aligned} |z^2+2z-5| & \le |z^2|+2|z|+5 \\ &\le(1+2+5)|z|^2\ , \\ |z^2+4| & \ge |z^2|-4 \\ &\ge|z|^2-\frac 4{100}|z^2|=\left(1-\frac 4{100}\right)|z^2|\ , \\ |z^2+2z+2| & \ge |z^2|-2|z|-2 \\ &\ge|z|^2-\frac 2{10}|z^2|-\frac 2{100}|z^2| =\left(1-\frac 2{10}-\frac 2{100}\right)|z^2|\ , \end{aligned} $$ so the whole integrand can be controlled, bounded by a constant times $|z^{-2}|=r^{-2}$. This times the length of the contour, $2\pi r$, tends to zero.

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To use the residues, it is enough to observe that the sum of the residues (inside of a circle around zero containing all for poles) is zero. So the integral is zero. (Also after considering the $2\pi i$ from the formula.)

Let us compute the residues.

To makes things easy (to type), i will use a computer algebra system, here sage:

sage: var( 'z' );
sage: F = (z^2+2*z-5)/(z^2+4)/(z^2+2*z+2)

sage: F.partial_fraction()
1/10*(7*z + 17)/(z^2 + 4) - 7/10*(z + 3)/(z^2 + 2*z + 2)

sage: poles = F.denominator().roots(multiplicities=False)
sage: poles
[-I - 1, I - 1, -2*I, 2*I]

sage: for pole in poles:
....:     print "Residue of F in %6s is %15s" % ( pole, F.residue(z==pole) )
....:     
Residue of F in -I - 1 is  -7/10*I - 7/20
Residue of F in  I - 1 is   7/10*I - 7/20
Residue of F in   -2*I is  17/40*I + 7/20
Residue of F in    2*I is -17/40*I + 7/20

From the above, it is clear that the sum of the residues is zero, the terms in $\pm 7/20$, in $\pm 7i/10$, and in $\pm 17i/10$ cancel together.

We use the above partial fraction decomposition to make the computations easy. The residue of $$ \frac{7 \, z + 17}{10 \, {\left(z^{2} + 4\right)}} = \frac 1{z-2i}\cdot\frac{7 \, z + 17}{10 \, {\left(z+2i\right)}} $$ in $2i$ is the value of the function $$ \frac{7 \, z + 17}{10 \, {\left(z+2i\right)}} $$ in $2i$, we have only to substitute $z=2i$ and compute $\frac{14i+17}{10\cdot(2i+2i)}$ explicitly. We get the above value in the line Residue of F in 2*I is -17/40*I + 7/20. The residue in $-2i$ is the complex conjugate. (Since the fraction is invariated by the complex conjugation.) Similarly we compute the residues for the other poles and the other fraction in the partial fraction decomposition...

Answer 2

Can't you re-write the integral using partial fractions as the following

$$\frac{1}{10} \left[ \int \frac{7z+17}{z^2 + 4} dz + \int \frac{7(z+3)}{z^2 + 2z +2} dz \right].$$

Both denominators can be expressed as $z^2 + 4 = (z+2i)(z-2i)$ and similarly $z^2 + 2z + 2 = (z-z_1) (z-z_2)$ where $z_1 = -1+i$ and $z_2 = -1-i$. Then follow the instructions from the wiki article.

Answer 3

I hope the idea of residue at infinity would delight you.

For a complex function $f(z)$, its residue at infinity is defined as $$\text{Res}_{z=\infty}f(z)=-\text{Res}_{z=0}\frac1{z^2}f(\frac1z)$$

A useful relation is $$\text{Res}_{\infty}f(z)+\sum_{\text{all poles}} \text{Res}\, f(z)=0$$ for $f(z)$ with finitely many poles.

Since your integrand has finitely many poles, the formula applies.

Let $f(z)$ be your integrand.

With a little algebra, we get $$\text{Res}_\infty f(z)=-\text{Res}_{z=0}\frac{1+2z-5z^2}{(1+4z^2)(1+2z+2z^2)}\color{RED}{=0}$$

Immediately, $$\sum_{\text{all poles}} \text{Res}\, f(z)\color{RED}{=0} $$ as desired.

Please also note that, due to Cauchy’s theorem, $$\oint_{|z|=\infty}f(z)dz=\sum_{\text{all poles}}\text{Res}f(z)=\color{red}{0}$$ and your result follows. Plus, thus the tedious calculation of residues can be avoided.