Complex integral $\oint_L \frac{\cos^2{z}}{z^2}dz$

Question

Compute $$ \int_L \frac{\cos^2 z}{z^2}\,dz$$ where $L$ is the closed loop that goes counterclockwise around the square with vertices $-1$, $-i$, $1$ and $i$.

I was trying to compute this using the series expansion. I had that $$\cos^2{z}=1-z^2+\frac{z^4}{3}-\frac{2z^6}{45}+\cdots,$$ and so $$\oint_L\frac{\cos^2{z}}{z^2}dz=\frac{1}{z^2}-1+\frac{z^2}{3}-\frac{2z^4}{45}+\cdots$$

I was hoping to be able to say that all but 1 of these integrate to zero, which I believe is the case. But the term that doesn't integrate to zero is -1. This makes the total integral $-1z\cdot 2\pi i \cdot v(L,0)$. Which I believe is $-2\pi i$.

I think that I have done something wrong.

Help?

Answer 1

By Cauchy formula, $$\int_L\frac{\cos^2 z}{z^2}dz=2\pi if'(0)=0$$

where $f(z)=\cos^2 z$.

An equivalente way to do is to use the residue theorem. As you said $$\frac{\cos^2 z}{z^2}=\frac{1}{z^2}-1+...$$ therefore $$\text{Res}_{z=0}\left(\frac{\cos^2 z}{z^2}\right)=0$$

and thus,

$$\int_L\frac{\cos^2 z}{z^2}dz=2\pi i\text{Res}_{z=0}\left(\frac{\cos^2 z}{z^2}\right)=0.$$

Answer 2

We can easily see that $z_{0}$ is an isolated singular point because it is the only singular in the given contour $L$ so,

$$\int\limits_{L} \frac{cos^2 z}{z^2}= 2\pi i ( \substack{Res}_{z=0} f(z)) $$

where $f(z)=\frac{cos^2 z}{z^2}$

What is residue of $f(z)$ at $z=0$?

Hint:The coefficient of $(z-0)^{-1}$ in the expansion of $f(z)$