Evaluate $\int_{|z-1|=3} \frac{z-2}{z(z-1)(z+2)}dz$ using residue theorem

Question

I need to evaluate $$\int_{|z-1|=3} \frac{z-2}{z(z-1)(z+2)}dz$$ So the integrand has 3 poles of order 1: $z=0$, $z=1$ and $z=-2$. But $z=-2$ lies on the contour itself. I found out that I need to calculate Cauchy Principal Value, but I can't understand how.

Answer 1

The principal value is normally assessed by cutting out a small portion, say length $2 \varepsilon$, symmetrically on either side of the singularity, evaluate the integral on the resulting contour and then allow $\varepsilon \to 0$. If this has a limit then that is the principal value.

Applying it to your case, let $I$ be the integral taken round the circular contour but stopping at a distance of $\varepsilon$ each side of $z=-2$. Then add each of two paths to $I$. The first, $J_1$ joins the ends of $I$ in a nearly circular arc so that combined $I$ and $J_1$ enclose $z=-2$. The second, $J_2$, also joins the ends of $I$ using a circular arc, but excluding $z=-2$.

If the residues at the poles are $r_1, r_2, r_3$, with that at $z=-2 = r_1$ then, irrespective of $\varepsilon$, $$I+J_1 = 2\pi i (r_1+r_2+r_3), \quad I+J_2 = 2\pi i(r_2+r_3)$$ Now the "Indentation Lemma" states that if $f$ is holomorphic except at a simple pole at $z=a$ where it has residue $b$, taking a contour $\gamma=a + \varepsilon e^{i\theta}$ for $\theta_1 \leqslant \theta \leqslant \theta_2$, then as $\varepsilon \to 0$, $$\int_{\gamma} f(z)dz \to ib(\theta_2-\theta_1).$$

Using it for $J_1$ and $J_2$, noting their respective orientations, as $\varepsilon \to 0$ both $J_1, J_2$ become semi-circular, and by this lemma, $J_1$ and $-J_2$ will have limit $\pi i r_1 $ and $J_1+J_2 \to 0$. We now have, $$2I = 2 \cdot 2\pi i (r_2+r_3) + 2\pi i r_1 - (J_1+J_2).$$ Let $\varepsilon \to 0$, to obtain, $$I = 2\pi i \big(\tfrac{1}{2} r_1+r_2+r_3 \big)$$