I'm taking my first course in complex analysis and I've come across a question I haven't encountered before.
$$\int_C \frac{z dz}{(z+2)(z-1)}$$
Such that C is a circle where $|Z| = 4$.
So my approach was,
seeing that this has singularities, $z = -2, 1$. I see that they lie within the circle. Therefore, I used partial fractions to break this up into two integrals.
$$\int_C \frac{z}{3(z-1)}dz – \int_C \frac{z}{3(z+2)}dz$$
Then applying cauchy's integral formula on these seperate integrals I get. Where $f(z) = z$ in both cases.
$$= \frac{2}{3} \pi i f(1) – \frac{2}{3}\pi if(-2)$$
$$ =\frac{2}{3} \pi i + \frac{4}{3} \pi i $$
$$= 2\pi i $$
Is this the correct way to approach this integral? Thank you for any guidance.
Best Answer
Using partial fractions, $$ \frac{z}{(z+2)(z-1)}=\frac{2}{3(z+2)}+\frac{1}{3(z-1)} $$ Therefore, taking the contour integral over the curve (|z|=4), $$ \oint_{|z|=4}\frac{z}{(z+2)(z-1)}\,dz=\oint_{|z|=4}\left(\frac{2}{3(z+2)}+\frac{1}{3(z-1)}\right)\,dz $$ By linearity of integration, this becomes $$ \oint_{|z|=4}\left(\frac{2}{3(z+2)}+\frac{1}{3(z-1)}\right)\,dz=\oint_{|z|=4}\left(\frac{2}{3(z+2)}\right)\,dz+\oint_{|z|=4}\left(\frac{1}{3(z-1)}\right)\,dz $$ Cauchy's integral formula states that for a suitable function $f(z)$ and curve $\gamma$, $$ f(a)=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-a}\,dz $$ where $a$ is any point within the interior of the curve $\gamma$. In this case, the function $f(z)$ is constant, and the singularities are $z=-2$ and $z=1$ are within the interior of the curve, so applying the formula gives $$ \begin{align*} \oint_{|z|=4}\left(\frac{2}{3(z+2)}\right)\,dz+\oint_{|z|=4}\left(\frac{1}{3(z-1)}\right)\,dz&=\frac{4\pi i}{3}+\frac{2\pi i}{3}\\ &=2\pi i. \end{align*} $$