How to evaluate the following complex integral

Question

I'm asked to evaluate the following complex integral

$$\oint_{C}^{}{\frac{z}{z^2+a^2}dz}\quad |a|<1$$

where C is the unit circle enclosing the origin.

The integrand loses analyticity when $z = \pm ia$, each of which lies interior to C.

However, we can utilize deformation of the contour and Cauchy's Theorem where $C=C_0-C_1-C_2=0$. Therefore we now need to evaluate

$$\oint_{C_1}^{}{\frac{z}{z^2+a^2}dz} + \oint_{C_2}^{}{\frac{z}{z^2+a^2}dz}\quad |a|<1$$

However, I don't know how to evaluate the above integrals, and would love some advice on how to do so.

Answer 1

METHODOLOGY $1$:

Yet another approach is to use partial fraction expansion and write

$$\frac{z}{z^2+a^2}=\frac12\left(\frac{1}{z-ia}+\frac{1}{z+ia}\right)$$

Then, we have

$$\begin{align} \int_{|z|=1}\frac{z}{z^2+a^2}\,dz&=\frac12\int_{|z|=1}\left(\frac{1}{z-ia}+\frac{1}{z+ia}\right)\,dz\\\\ &=\frac12 \int_{|z|=1}\frac{1}{z-ia}\,dz+\frac12 \int_{|z|=1}\frac{1}{z+ia}\,dz\\\\ &=\frac12 \left(2\pi i+2\pi i\right)\\\\ &=2\pi i \end{align}$$

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METHODOLOGY $2$:

We can evaluate the integral of interest by evaluating the Residue at Infinity. Let $f(z)=\frac{z}{z^2+a^2}$, then the residue at infinity of $f$ is given by

$$\begin{align} \text{Res}\left(f(z),\infty\right)&=-\text{Res}\left(\frac{1}{z^2}f\left(\frac1z\right),0\right)\\\\ &=-\lim_{z\to 0}\left(\frac{1}{a^2z^2+1}\right)\\\\ &=-1 \end{align}$$

Therefore, we find that

$$\begin{align} \int_{|z|=1}f(z)\,dz&=-2\pi i\, \text{Res}\left(f(z),\infty\right)\\\\ &=2\pi i \end{align}$$

as expected!

Answer 2

Using the factorization given by Daniel in the comments and Cauchy's Integral Theorem the integrals evaluate to

$$2\pi i(\frac{ia}{2ia})+2\pi i(\frac{-ia}{-2ia}) = 2\pi i$$