Lecture Notes

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====== September 20 (Wednesday) ======

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 $f(z) =  \frac{1}{(z-1)(z-2)}$. 

===== Method 1: residue theorem =====
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 
$$ I = (2 \pi i) [Res_{z=1} f(z) + Res_{z=2} f(z)] = 0 .$$

===== Method 2: make the radius $R$ large =====
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 $I_R=0$. 

===== Method 3: change of variable =====
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. 

===== Riemann sphere =====
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 $w=0$.



===== Exercises =====
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)
Lecture Notes

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