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 $I_R$ is a constant, and for any $R>10$, we have $0 \leq |I_R| \leq \frac{2\pi R}{(R-1)(R-2)}.$ Thus $|I_R|=0$, hence $I_R=0$.

Let $w = 1/z$, then we have

$I = \oint_{|w|=1/10, CW} \frac{1}{(w^{-1}-1)(w^{-1}-2)} d (w^{-1}) = \oint_{|w|=1/10, CW} \frac{-1}{(1-w)(1-2w)} dw = \oint_{|w|=1/10} \frac{1}{(1-w)(1-2w)} dw$ where in the last step, I changed the orientation of the contour from $CW$ to $CCW$(CCW is by default, hence omited) and add an extra $(-1)$ factor to the integral.

Since the integrand is only singular at $w=1,1/2$, and the contour $|w|=1/10$ contains no singularity in its interior, the integral is 0.

It is useful to think of add a point $\infty$ to the complex plane $\C$, and think of $\C \cup \{\infty\}$ as a sphere, where $\infty$ is identified with the north pole, $0$ with the south pole, the unit circle $|z|=1$ as the equator.

The natural coordinate to use near the north pole is $w=1/z$, so that $z=\infty$ corresponds to $w=0$.

Let $C$ be the contour of $|z|=10$. Consider the following integrals.

(1) $\oint_C \frac{1}{1+z^2} dz$

(2) (the result for this one is not zero.) $\oint_C \frac{z}{1+z^2} dz$

(3) $\oint_C \frac{z^2}{1+z^4} dz$

Apply methods 1,2,3 to the above problems (each method need to be used once)