Today we discussed this integral $$ \oint_{|z|=10} \frac{1}{(z-1)(z-2)} dz $$ where the contour is a CCW (counter-clockwise) circle of radius $10$ (or any radius $R > 2$, that encloses $1,2$)

We used three methods to show that this is zero. Denote the integrand by $f(z)$, namely $f(z) = \frac{1}{(z-1)(z-2)}$.

We apply residue theorem, and computed $$ Res_{z=1} f(z) = \frac{1}{z-2}|_{z=1} = -1. $$ $$ Res_{z=2} f(z) = \frac{1}{z-1}|_{z=2} = 1. $$ Hence the result of the contour integral, denoted by $I$ is $$ I = (2 \pi i) [Res_{z=1} f(z) + Res_{z=2} f(z)] = 0 .$$

We can deform the contour $C$, as long as the integrand $f(z)$ when restricted to the contour remains a holomorphic function during the deformation. The resulting integral is invariant under the deformation.

Thus, we may change the circle $|z|=10$ to $|z|=R$, and let $R$ tends to $\infty$. Let $I_R$ be the integral on the contour $|z|=R$. We have two claims

• $I_R$ is independent of $R$. This is because the function $f(z)$ has no pole in the region swept out by the contour deformation, namely between $|z|=10$ and $|z|=R$.
• We have the following estimate of $|I_R|$.

$$ |I_R| \leq \oint_{|z|=R} | f(z) | |dz| \leq \oint_{|z|=R} \frac{1}{|z-1| |z-2| } |dz| $$ By triangle inequality for complex numbers, $$ |a| + |b| \geq |a+b| \geq | |a| - |b| |, $$ we have for any $z$ with $|z|=R > 10$ $$ |z-1| \geq |z|-1 = R-1 > 0, \quad |z-2| \geq |z|-2 = R - 2 > 0, $$ hence $$ \frac{1}{|z-1| |z-2| } \leq \frac{1}{(R-1)(R-2)} $$ on the contour $|z|=R$. Thus $$ \oint_{|z|=R} \frac{1}{|z-1| |z-2| } |dz| \leq \frac{1}{(R-1)(R-2)} \oint |dz| = \frac{1}{(R-1)(R-2)} (2\pi R) $$ To summarize, we have $$ 0 \leq |I_R| \leq \frac{2\pi R}{(R-1)(R-2)}. $$