How to compute the $\int_C \frac{ z}{(z+2)(z-1)} dz$ where $c = \{|z| = 4\}$ is the circle of radius $4$ centred at zero oriented anticlockwise

Question

First of all, how to break up the function $$\frac{z}{(z+2)(z-1)}$$ to be able to perform the integral? Then I think I know how to compute it, any help will be greatly appreciated.

Answer 1

By partial fractions, your integral is equivalent to

$$\int_c \frac{2 dz}{z + 2} - \int_c \frac{dz}{z + 1}$$

The first integrand has a pole at $z = -2$, which is inside $c$, and has an easily computed residue. Likewise for the second, which has a pole at $1$. Now use the residue theorem.

Answer 2

Given you have a circle of radius $4$, you see immediately that your points of singularity are all included, namely $z=-2,1$.

1. By the Residue Theorem $\int_c \frac{z}{(z+2)(z-1)}=2\pi i$(sum of the residues for $f$ at $z=1,-2$).

2. Well if we wanted to look at the residue of $f$ at z=1 we have $\frac{\frac{z}{z+2}}{z-1}=\frac{a_0+a_1z+a_1z^2+...}{z-1}$, but since $\frac{z}{z+2}$ is analytic at $z=1$, it is equal to it's Taylor series at this value. Thus, the res $f(z)$ at $z=1$ is, $\frac{1}{1+2}=\frac{1}{3}.$

3. Similarly for the res $f(z)$ at $z=-2$ we have,$\frac{\frac{z}{z-1}}{z+2}=\frac{a_0+a_1z+a_1z^2+...}{z+2}$. Again, $\frac{z}{z-1}$ is analytic at $z=-2$, it is equal to it's Taylor series at this value. Thus, the res $f(z)$ at $z=-2$ is, $\frac{-2}{-2-1}=\frac{2}{3}.$

4. Hence by the Residue Theorem, $\int_c \frac{z}{(z+2)(z-1)}=2\pi i(\frac{2}{3}+\frac{1}{3}) = 2\pi i.$ And we didn't have to do partial fractions. :)