Complex contour integral $\int_C \dfrac{cosh(z)}{z^4}dz$ where C is the square centered at the origin with side length 4

Question

Can somebody explain how to carry out the integral $\int_C \dfrac{cosh(z)}{z^4}dz$ where C is a square centered at the origin with side length 4? I know I'm supposed to use the Cauchy integral formula, but when I carry out the integral I get $2\pi i/6$. My textbook gives the answer as $0$.

Answer 1

$f(z)=\frac{\cosh z}{z^4}$ is a meromorphic function with a single pole of order four at the origin.
By the residue theorem, if $\gamma$ is any closed simple curve enclosing the origin, counter-clockwise oriented, $$ \oint_{\gamma} f(z)\,dz = 2\pi i\cdot\text{Res}\left(f(z),z=0\right). $$ However, $\cosh(z)$ is an entire and even function, so a holomorphic function of $z^2$, and the Laurent expansion at the origin of $\frac{\cosh z}{z^4}$ contains no term of the form $\frac{1}{z}$. In particular, $$\text{Res}\left(f(z),z=0\right)=\color{red}{\large 0}.$$