Find $\int_{|z|=\frac 32}\frac{2z-3}{z^2-3z+2}dz$ complex integral

Question

So to solve the integral $$\int_{|z|=\frac 32}\frac{2z-3}{z^2-3z+2}dz$$ I have 2 ideas but I don't know which one to use or if both are valid or not, one of them is to find the residue and solve it with the formula $$\int_\gamma f(z)dz=2\pi i \text{Res}(f,z_k)$$ or if I can just separate it with partial fractions such as so $$\int_{|z|=\frac 32}\frac{2z-3}{z^2-3z+2}dz=\int_{|z|=\frac 32} (-\frac{1}{1-z}-\frac{1}{2-z})dz$$ and use the Cauchy formula for integrals $$\int_\gamma \frac{f(z)}{z-z_0}dz=2\pi i f(z_0)$$ would both method be valid? Or would the last one not be valid due to the ceros of the function?

Answer 1

So the first method (residue theorem) is definitively valid (and yields $2\pi i$).

For the second one, we can do partial fractions as you suggested, which yields $$\oint_{|z|=\frac32}\frac{2z-3}{z^2-3z+2}dz = \oint_{|z|=\frac32}\frac{1}{z-1} dz+ \oint_{|z|=\frac32}\frac{1}{z-2}dz$$ The second integral's pole doesn't lie in the contour so it goes to $0$. The first integral is $2\pi i$ by the Cauchy integration formula, with $a=1$ and $f(z)=1$ as seen in the statement of theorem, so the second method is definetely also valid.

Answer 2

1. Expand the denominator into a product of monomials. The zero of each of the monomials is a pole, if it does not cancel out with a zero of the numerator at the same point. What is the degree of each of the poles?

2. Calculate the residue of the function at each of the poles. Hint: what is the residue of a function of the form $g(z)/(z-z_0)$ at $z_0$, where $g(z)$ is analytic in the neighborhood of $z_0$?

3. Use the residue theorem. Indicate the poles inside your integration contour. Sum the residues inside the contour, multiply by $2\pi i$. By the residue theorem, this is your integral.

Answer 3

The integrands singularities are $z_1=1$ and $z_2=2$. Thus, since the contour only includes $z_1$, you can either use residue theorem for $z_1$, Laurent series, or Cauchy's integral formula, which are all similar.

$\mathbf{Residue's\ Theorem:}$ $$I=\int_{\gamma:|z|=\frac{3}{2}}\dfrac{2z-3}{z^2-3z+2}dz=2\pi i\sum_{\{k|z_k\in\text{Int}(\gamma)\}}\text{Res}(f(z),z_k)=2\pi i\text{Res}(f(z),z_1)$$ By the definition and knowing that $z_1$ is a $1^{\text{st}}$ order pole, $$\text{Res}(f(z),z_1)=\lim_{z\to 1}\left[(z-1)\dfrac{2z-3}{(z-1)(z-2)}\right]=1.$$

Therefore, $I=2\pi i$.

$\mathbf{Laurent\ Series:}$

$$ \begin{aligned} f(z)=\dfrac{2z-3}{z^2-3z+2} &=\dfrac{2z-3}{(z-1)(z-2)}\\ &=-\dfrac{2z-3}{z-1}\dfrac{1}{1-(z-1)}\\ &=\dfrac{1-2(z-1)}{z-1}\sum_{n\geq0}(z-1)^n\\ &=\left(\dfrac{1}{z-1}-2\right)(1+(z-1)+(z-1)^2+...)=\dfrac{1}{z-1}+...\implies a_{-1}=1. \end{aligned} $$ Since $\text{Res}(f(z),z_k)=a_{-1}$ of $f(z)$ developed around $z_k$, we get that $$I=2\pi i\text{Res}(f(z),z_1)=2\pi ia_{-1}=2\pi i.$$

$\mathbf{Cauchy's\ Formula:}$

The statement is $$\int_{C}\dfrac{f(z)}{z-z_o}dz=2\pi if(z_o),\ z_o\in\text{Int}(C).$$ Since the only point inside $\gamma$ is $z_1$, $z_o=z_1$ and $f(z)=\frac{2z-3}{z-2}$: $$I=\int_{\gamma:|z|=\frac{3}{2}}\dfrac{2z-3}{z^2-3z+2}dz=\int_{\gamma}\dfrac{\frac{2z-3}{z-2}}{z-1}=2\pi i\left(\frac{2z-3}{z-2}\right|_{z=1}=2\pi i.$$

$\mathbf{Proof\ of\ Res(...)=a_{-1}:}$

Differentiating Cauchy's formula wrt the parameter $z_o$ $k$ times, you can get that $$k!\int\dfrac{f(z)}{(z-z_o)^{k+1}}dz=2\pi if^{(k)}(z_o).$$ Taylor series' coefficients are defined like $a_k\equiv f^{(k)}(z_o)/k!$, therefore, $$\int\dfrac{f(z)}{(z-z_o)^{k+1}}dz=2\pi ia_k.$$ Finally, setting $k=-1$, we get that $$\int f(z)dz=2\pi i a_{-1}\equiv2\pi i\text{Res}(f(z),z_o).$$

Answer 4

You're making it too difficult. You have singularities at $z=1$ as $z=2$ lies outside the circle, so your result is

\begin{align} \int_{\vert z \vert = \frac{3}{2}} f(z)dz&=2 \pi i\bigg(\frac{2z-3}{z-2}\bigg\vert_{z=1}\bigg)\\ &=2\pi i (1)\\ &=2 \pi i \end{align}