Residues of Complex Functions

Question

I need to find the residues of $f$ at the isolated singular points, namely $z=1,z=0$.

Where $f(z)=\dfrac{2z+1}{z(z+1)}$.

I already have that the residue at $z=0$ is $1$, and I know I need to do some slight of hand to get res at $z=1$.

I tried, $f(z)=\dfrac{2z+1}{z(z+1)} = (\dfrac{2z+1}{z})(\dfrac{1}{2+(z-1)}) = (2+d\dfrac{1}{z})*\sum_{n=0}^{\infty}(-1)^n\dfrac{(z-1)^n}{2^{n+1}}$, but we never get a $\dfrac{1}{z-1}$ term, so can we conclude that the coefficient of $\dfrac{1}{z-1} = res =0?$

I would need this to conclude that, $\int_C f(z) dz = 2i\pi(1+0)=2i\pi$. Where C is the circle of radius 2.

Yes, I know I can use partial fractions to do this, but our professor wants us to use the Residue Theorem to evaluate the integral.

Answer 1

Hint:

$$\frac{2z+1}{z(z+1)}=\frac1z+\frac1{z+1}=-\frac1{1-(z+1)}+\frac1{z+1}=$$

$$-1-(z+1)-(z+1)^2-\ldots+\frac1{z+1}\;\;\implies\;\ldots$$

Answer 2

Given $$f(z)=g(z)/(z-z_0)$$ where $f(z)$ is analytic at $z_0$ the residue of $g(z)$ at $z_0$ is simply $g(z_0)$. So the residue at $z=-1$ is given by $\frac{2z+1}{z}$ evaluated at $z=-1$ which gives a value of $1$.

As an extension of this, you may be interested in the Cauchy Integral Formula.