How do I evaluate $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$

Question

How do I evaluate the following integral when where $C$ is the square with vertices at $\pm2, \pm2+4i$,

$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$$

Using Cauchy integral:

$\frac{z^{2}}{z^{2}+4}=\frac{z^2}{(z+2i)(z-2i)}=\frac12\frac{z^2}{z+2i}+\frac12\frac{z^2}{z-2i}$ then $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4} \implies \frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}$

$\frac{z^2}{z+2i}$ is analytic on and inside $C$, hence we can apply Cauchy theorem and for the second term we use Cauchy integral formula,

$$\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}=0+\frac{1}{4\pi i}2\pi i \times f(2i)=-2$$

Using Residue Theorem:

$2i$ is the only isolated singularity in $C$.

$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}=\frac{1}{2 \pi i} 2\pi i \times \text{Res}(f,2i)=i$$

I get correct answer for Residue Theorem but couldn't understand where I do wrong when using Cauchy integral.

It will be great help if someone clear me when to use which method to find the integral.

Answer 1

Here you don't have to use partial fractions but use $f(z) = \frac{z^2}{z+2i}$ for $f(a)$ in the Cauchy Integral Formula, since $f$ is holomorphic in the smallest open disk that fits your $C$.

Then you get $f(2i) = \frac{(2i)^2}{4i} = i$

Answer 2

I will perform the first computation in a slightly different manner: $$ \begin{aligned} \frac{1}{2 \pi i} \oint_{C} \frac{z^2 }{z^2+4}\; dz &= \frac{1}{2 \pi i} \oint_{C} \frac{(z^2 + 4) - 4}{z^2+4}\; dz \\ &= \frac{1}{2 \pi i} \oint_{C} dz + \frac{1}{2 \pi i} (-4)\oint_{C} \frac1{z^2+4}\; dz \\ &= 0 + \frac{1}{2 \pi i} (-4)\oint_{C} \frac 1{4i}\left(\frac1{z-2i} - \frac1{z+2i}\right)\; dz \\ &= \frac{1}{2 \pi i} \cdot i\oint_{C} \frac{dz}{z-2i} - \frac{1}{2 \pi i} \cdot i\oint_{C} \frac{dz}{z+2i} \\ &= i-0\ . \end{aligned} $$