Table of Contents
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)
September 18: Residue theorem
Let $\Omega$ be a domain (open subset), and $z_0 \in \Omega$ a point. We say a holomorphic function $f: \Omega \RM \{z_0\} \to \C$ has a pole of order $m$ at $z_0$, if $h(z) = (z-z_0)^m f(z)$ can be extended to the entire $\Omega$, and $h(z_0) \neq 0$.
Suppose $f$ has an order $m$ pole at $z_0$, then near $z_0$ (more precisely, for $z \in \C$ with $0 < |z-z_0| < r$ for some small positive $r$), we can write Laurent expansion $$ f(z) = \frac{a_{-m}}{ (z-z_0)^m} + \cdots + \frac{a_{-1}}{ (z-z_0)^1} + a_0 + a_1 (z - z_0)^1 + \cdots $$
The coefficient $a_{-1}$ is called the residue, and denoted as $Res_{z_0} f := a_{-1}$. ($:=$ read 'is defined as')
How to find the residue ?
1. If the pole is order $1$, then $Res_{z_0}f(z) = \lim_{z \to z_0} f(z) (z-z_0)$.
Example: residue of $f(z) = (z-2) / (z-1)$ at $z=1$.
2. If the pole at $z_0$ is of order $m>1$, then just Laurent expand. More precisely, do Taylor expansion of $$(z-z_0)^m f(z) = a_{-m} + \cdots + a_{-1} (z-z_0)^{m-1} + \cdots $$ and then you get $a_{-1}$.
Example: residue of $f(z) = [(z-2) / (z-1)]^2$ at $z=1$.
Residue Theorem
Thm: let $f$ has a pole (of some order) at $z_0$, and let $C$ be a small circle of radius $\epsilon$ around $z_0$ positively oriented (i.e in counter-clockwise direction) so that $f$ is holomorphic on $C$, and $C$ encloses only one pole. Then $$\oint_{C} f(z) dz = (2\pi i) Res_{z_0} f. $$
Proof sketch: write $f$ as a Laurent expansion, then integrate term by term.
Homework 4
Due Monday in class.
1. Taylor expand $(z+1)(z+2)$ around $z=3$.
2. Laurent expand $1/[(z-1)(z-2)]$ around $z=1$. And do it again, this time around $z=2$.
3.Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $ for the following three contours (a) For $t \in [0, 2\pi]$, let $z(t) = e^{it}$.
(b). For $t \in [0, 2\pi]$, let $z(t) = e^{i2t}$.
(c ). For $t \in [0, 2\pi]$, let $z(t) = e^{-it}$.
Boas, Ch 14, Section 3, #4, 6
Contour Integral: exercise day
Let's see some cool application of contour integral first.
1. $\int_\R 1/(1+x^2) dx$
didn't finish, the rest will wait till next week.
2. $\int_0^\infty 1/(1+x^3) dx$
3. $\int_0^\infty e^{-x}/(1+x^3) dx$
4. $\int_\R \frac{x}{x^4+1} e^{-x^2} dx$
Contour Integral
Reading: Boas, Ch14, section 1-5
So, you have learned what holomorphic function looks like, and you know there are functions which are 'bad' only at a few points. What do you want to do with these functions?
primitive of a holomorphic function on $\C$
Just like calculus, you can do differentiation, and you can do integration. Differentiation is easy, let's talk about integration.
Recall what we do in real analysis case: given $f(x)$ on $\R$, we can find one primitive $F(x)$ by considering $$ F(x) = C + \int_{x_0}^x f(u) du $$ where we set the initial condition that $F(x_0) = C$, and $F'(x)= f(x)$.
Can we do the same here? Say $f(z)$ is a holomorphic function, we can define $$ F(z) = C + \int_{z_0}^z f(u) du $$ Now, we immediately run into trouble: how do we go from $z_0$ to $z$? Does the integration depends on how we choose the path from $z_0$ to $z$? Thanks to the fact that $f$ is holomorphic, the integration is independent of the choice of path.
primitive of $1/z$
OK, $f(z) = 1/z$ is not a holomorphic function on the entire $\C$. We can say, it is a holomorphic function on the 'punctured complex plane' $\C^* = \C \RM \{0\}$, or it is a meromorphic function on $\C$ with a pole of order $1$ at $z=0$. Either way, we can ask, can we find the primitive of $1/z$ on $\C^*$? Namely, is there a hol'c function $F(z)$ such that $F'(z) = 1/z$?
You probably know that, for $x>0$, if you integrate $1/x$, you get $\log (x) + C$. (why is that? )
The same holds for complex analytic function. almost. We can say the primitive of $1/z$ is $\log(z) + C$, but $\log(z)$ is a multivalued function on $\C^*$.
Contour integral for meromorphic function
Let $f: \Omega \to \C$ be a meromorphic function. Let $\gamma$ be a closed contour in $\Omega$ (“contour” just means a smooth path) that avoids the pole of $f$. Then, $$ \int_\gamma f(z) dz = (2 \pi i) \sum_{z_0 \text{poles of } f} Res_{z_0} f. $$ The $Res_{z_0} f$ is the residue of $f$, which is the Laurent expansion of $f$ at $z$, the coefficient in front of $1/(z-z_0)$.
Example: $f(z) = 1/ [(z-1)(z-2)(z-3)]$, $\gamma$ is a contour around the two poles $1$ and $2$.
Exercise
1. For $t \in [0, 2\pi]$, let $z(t) = e^{it}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
2. For $t \in [0, 2\pi]$, let $z(t) = e^{i2t}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
3. For $t \in [0, 2\pi]$, let $z(t) = e^{-it}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
