# Contour Integation $\int_\gamma \frac{\cos^2z}{z^2}$

I have the following question from a past exam paper that I'm not really sure how to evaluate. Any help would be appreciated...

Let $\gamma$ be the unit circle in $\mathbb{C}$ traversed in the anti-clockewise direction.

$$\displaystyle\int_\gamma \dfrac{\cos^2z}{z^2}dz$$

I know that $\gamma(t)=e^{it}$, as it is the unit circle with a center 0 and radius 1.

• Hint: $\cos z=1+z^2/2+\ldots$ whence $\cos^2z=1+z^2+\ldots$. Which coefficient does this integral compute, up to multiplication by $2i\pi$? Commented May 1, 2013 at 17:18

We can use Cauchy's Integral Formula, this tells that $2 \pi i(\cos^2(0))'=\int_\gamma \dfrac{\cos^2z}{(z-0)^2}=0$.
One way to look at this is to see that $\cos^2{z} = 1 - \sin^2{z}$, and that
$$\oint_{\gamma} dz \frac{\cos^2{z}}{z^2} = \oint_{\gamma} dz \frac{1}{z^2} - \oint_{\gamma} dz \frac{\sin^2{z}}{z^2}$$
The first integral on the RHS is zero; you may see this immediately by substituting $z=e^{i \theta}$ into the integral. The second is zero by Cauchy's integral theorem, because $\sin{z}/z$ is analytic at $z=0$. Thus, the integral is zero.