Complex Integral: $\int_{|z| = 1}\frac{z^2}{4e^z -z}\,\mathrm dz$

Question

I am stuck computing the following complex integral $$\int_{|z| = 1}\frac{z^2}{4e^z -z}\,\mathrm dz$$

I do not even know if the integrating function has a pole and then using residue calculus.

Using the standard parametrization $z =e^{i\theta}$ does not seem to work as well.

Answer 1

For $|z| = 1$ we get $|4e^z-z - 4e^z| = |z| = 1 < 4/3 < 4e^{-1} \leq 4e^{Re(z)} = |4e^z|$. Rouché's theorem gives us that $4e^z-z$ has the same amount of zeros in $B(0,1)$ as $4e^z$ does. So your integrand doesnt have a pole in $B(0,1)$.

Hence, using Cauchy's integral theorem the Integral is $0$.