Complex Numbers II: A Guide to Contour Integration (Part 2 of 3)

by greenturtle3141, May 19, 2023, 5:09 AM

Lesson 5: Cauchy's Theorem

So here's a CRAZY fact.

Theorem (Cauchy): Suppose $f$ is a holomorphic function. Suppose $\gamma$ is a path that loops. If $f$ is defined everywhere inside the loop, then
$\int_\gamma f(z)\,dz = 0.$

Example 1

What is $\int_{\partial D} \sin(e^z) + e^{e^{e^{z}}} - e^{\cos(z^2)}\,dz?$ Remember that $\partial D$ is the unit circle oriented counter-clockwise. Well, the integrand is holomorphic, and it's defined on the entire plane (so it's defined inside the unit circle). Since the unit circle loops, the answer to this integral must be $\boxed{0}$ .

Example 2

What is $\int_\gamma \cos(\cos(\cos(\cos(\cos(z)))))\,dz$ , where $\gamma$ is the following curve?

$[asy] size(7cm); draw((-2,0)--(2,0),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-2)--(0,2),p=linewidth(1.5),arrow=Arrows(5)); draw((-1,-1)..(-1,0)..(-.5,1)..(-.2,-.2)..(0,0)..(1,1)..(0,.5)..(1.5,1)..(1.5,-.4)..(-1,-1),p=red+linewidth(1.75),arrow=MidArrow(7)); label("$\gamma$",(-1,1.5),p=red); [/asy]$

Well, the integrand is holomorphic, defined everywhere, and the path loops, so it's $\boxed{0}$ .

Example 3

Remember that annoying pacman integral $\int_\gamma e^z\,dz$ ?

$[asy] size(7cm); draw((-2,0)--(2,0),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-2)--(0,2),p=linewidth(1.5),arrow=Arrows(5)); draw((0,0)--dir(45)--arc((0,0),1,45,360-45,CCW)--(0,0),p=red,arrow=MidArrow); label("$\gamma$",(-1,1.5),p=red); [/asy]$

Well, $e^z$ is very holomorphic, is defined everywhere, and the path loops, so without needing to do any annoying computations I can say that the integral is $\boxed{0}$ .

Non-Example 1

It is not true that $\int_{\partial D} \overline{z}\,dz = 0$ . In fact, if we work it out, this integral will be equal to $2\pi i$ .

That's strange. $\partial D$ is just a circle, so it definitely loops... and $\overline{z}$ is defined inside the circle too. What went wrong? Can you figure it out? The reason is that...

Non-Example 2

It is not true that $\int_{\partial D} \frac{1}{z}\,dz=0$ . In fact, if we work it out, this integral will be equal to $2\pi i$ .

Huh? $\partial D$ is a looping path, and $\frac{1}{z}$ is holomorphic wherever it is defined. What went wrong this time? The reason is that...

Example 4

Let $f$ be a holomorphic function that is defined everywhere on $\mathbb{C}$ (i.e. " $f$ is entire"). Let $w_1$ and $w_2$ be two complex numbers, and consider two different paths that start at $w_1$ and end at $w_2$ . Call these paths $\alpha$ and $\beta$ .

$[asy] size(7cm); draw((-2,0)--(2,0),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-2)--(0,2),p=linewidth(1.5),arrow=Arrows(5)); draw((-1,-1)--(1,1),p=linewidth(1.5)+purple,arrow=MidArrow(5)); draw((-1,-1)..(-.5,1)..(0,1)..(1,1),p=linewidth(1.5)+red,arrow=MidArrow(5)); dot("$w_1$",(-1,-1),SW); dot("$w_2$",(1,1),NE); label("$\beta$",(-1,1.5),p=red); label("$\alpha$",(.5,-.5),p=purple); [/asy]$

It turns out that $\int_\alpha f(z)\,dz = \int_\beta f(z)\,dz$ ! This means that the path integral between $w_1$ and $w_2$ is the same, no matter what path you take.

Why is that the case? Let's reverse one of the paths... say, $\beta$ .

$[asy] size(7cm); draw((-2,0)--(2,0),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-2)--(0,2),p=linewidth(1.5),arrow=Arrows(5)); draw((-1,-1)--(1,1),p=linewidth(1.5)+purple,arrow=MidArrow(5)); draw((1,1)..(0,1)..(-.5,1)..(-1,-1),p=linewidth(1.5)+orange,arrow=MidArrow(5)); dot("$w_1$",(-1,-1),SW); dot("$w_2$",(1,1),NE); label("$-\beta$",(-1,1.5),p=orange); label("$\alpha$",(.5,-.5),p=purple); [/asy]$

Hey, now that's a loop! So by Cauchy's Theorem,
$\int_\alpha f(z)\,dz + \int_{-\beta} f(z)\,dz = 0.$ But remember: Reversing the direction of a path will negate the integral over it. So $\int_{-\beta} f(z)\,dz = -\int_{\beta} f(z)\,dz$ . This means that
$\int_\alpha f(z)\,dz - \int_{\beta} f(z)\,dz = 0.$ So yeah, the integrals are equal!

Exercises[/]

Compute the contour integral
$\int_{\partial D} e^{e^{e^{e^{e^{z}}}}}\,dz.$ Answer
It's just 0.
Remember Lesson 3 Exercise 1? You showed that for any $w \in \mathbb{C}$ that is not a negative real number, we have the identity
$\int_\gamma \frac{1}{z} \,dz = \log w,$ where $\gamma$ is some carefully-chosen path from $1$ to $w$ .

Prove that, actually, this identity holds no matter what path $\gamma$ we choose from $1$ to $w$ (as long as it doesn't cross the negative real line, where $\log$ would be undefined).

Solution
We use the principle that we learned from Example 4. If $\gamma$ is the carefully-chosen path from the exercise, and $\gamma'$ is some other path from $1$ to $w$ , then we want to show that the integrals $\int_\gamma 1/z\,dz$ and $\int_{\gamma'} 1/z\,dz$ are the same.

All we need to show that if we reverse the path $\gamma'$ to get $-\gamma'$ , then the paths $\gamma$ and $-\gamma'$ form a loop that we can apply Cauchy's Theorem to.

Well, $1/z$ is holomorphic over all $z \neq 0$ , so the only thing that can go wrong is if the singularity at $z=0$ is inside the loop formed by $\gamma$ and $-\gamma'$ . But this can't happen becaues neither of these two paths cross the negative real line!

So nothing can go wrong, and so we can apply Cauchy's Theorem to deduce that the two original integrals are indeed equal.

Lesson 6: Residues

The Residue Theorem is basically Cauchy's Theorem but better. To understand it, we need to know how to calculate a function's residue at a singularity.

I won't give you the "correct" definition of what a residue is, because that would make this post a lot longer and more complicated. So I'll just tell you how to calculate it in some simple cases.

General Setting: A holomorphic function $f$ may not be defined at some complex number $z_0$ . But, it would have something called a residue at this $z_0$ . A prototypical example is the function $1/z$ being not defined at $z=0$ . $1/z$ has no value at $z=0$ , but it has a residue.

General Idea: The residue of a function at a certain point tells you how much "spinning" it induces around that point.

How to calculate the residue in most cases: If $(z-z_0)f(z)$ is continuous at $z_0$ (meaning, we can "plug in the hole" at $z = z_0$ to make the function $(z-z_0)f(z)$ nice), then the residue is simply $\lim_{z \to z_0} (z-z_0)f(z)$ (i.e. the residue is the number we need to plug into the hole). The residue is denoted by $\operatorname*{Res}_{z=z_0} f(z)$ .

This is a very informal definition, so it will make a lot more sense when we try some examples.

Example 1

$1/z$ is not defined at $z=0$ , but it has a residue there. What is it? To find it, all we need to do is multiply by $(z-0) = z$ to make the singularity go away. Indeed, when we multiply $1/z$ by $z$ , we get $1$ , which is nice. The value of $1$ at $z=0$ is... well, $1$ . So the residue is $1$ . i.e. $\operatorname*{Res}_{z=0} \frac1z = 1$ .

Example 2

$\frac{1}{\sin z}$ is not defined at $z=0$ , but it has a residue there. What is it? To find it, we need to multiply by $(z-0) = z$ to try and make the singularity go away. When we multiply $1/\sin z$ by $z$ , we get $\frac{z}{\sin z}$ .

At first glance, it's not clear if we made the singularity go away. But if you took Calculus, you might know that $\lim_{x \to 0} \frac{x}{\sin x} = 1$ . The same limit holds in the complex world (and you can verify this by a Taylor expansion). That is, $\lim_{z \to 0} \frac{z}{\sin z} = 1$ . The $z$ factor helped make this limit possible! Because of this, the residue is easily found: It is exactly equal to that limit, which is $1$ . i.e. $\operatorname*{Res}_{z=0} \frac{1}{\sin z} = 1$ .

Example 3

$\frac{1}{z^2+1}$ is not defined at $z=i$ , but it has a residue there. What is it? To find it, we need to multiply by $(z-i)$ to make the singularity go away. Indeed,
$\frac{z-i}{z^2+1} = \frac{z-i}{(z-i)(z+i)} = \frac{1}{z+i}.$ This new expression has no singularity at $z=i$ anymore! So we successfully got rid of the badness at $z=i$ . The residue is then equal to the limit $\lim_{z \to i} \frac{1}{z+i}$ . But as I just said, this new function is nice at $z=i$ , so to evaluate the limit we can just plug in $z=i$ to get $\frac{1}{2i}$ . This is the residue! And we can write $\operatorname*{Res}_{z=i} \frac{1}{z^2+1} = \frac{1}{2i}$ .

Non-Example

$1/z^2$ is not defined at $z=0$ , but it has a residue there. What is it? To try and find it, we could multiply by $z-0 = z$ to get $1/z$ . But this doesn't remove the singularity because $1/z$ is still bad at $z=0$ . Oops!

That mean that the easy method I gave you doesn't work here. It turns out that $\operatorname*{Res}_{z=0} \frac{1}{z^2} = 0$ , for reasons that I will explain below.

Wait so what is the residue, actually? And how do I find the residue of higher-order poles, like in $1/z^2$ ?

To ease optional eyestrain, I'm going to put this in spoilers.

The Methodology and Rough Motivation

Exercises

Compute $\operatorname*{Res}_{z=42} \frac{\log z}{z - 42}$ .
Solution
Multiplying by $(z-42)$ removes the singularity at $42$ , leaving us with $\log z$ , and indeed this is nice at $z=42$ . Pluging in $z = 42$ gives $\boxed{\operatorname*{Res}_{z=42} \frac{\log z}{z - 42} = \log 42}$ .
Compute $\operatorname*{Res}_{z=i} \frac{1}{z^4-1}$ .
Solution
We write $\frac{1}{z^4-1} = \frac{1}{(z-1)(z+1)(z+i)(z-i)}$ . Now we remove the singularity by multiplying by $(z-i)$ . This leaves us with $\frac{1}{(z-1)(z+1)(z+i)}$ , which is now indeed nice at $z=i$ . Plugging in $z=i$ gives $\boxed{\operatorname*{Res}_{z=i} \frac{1}{z^4-1} = \frac{1}{(i-1)(i+1)(i+i)}}$ . Whatever that is.
(Using the Ugly Formula) Compute $\operatorname*{Res}_{z=1} \frac{\log z}{(z - 1)^3}$ .
Solution
First step: How many times do we need to multiply by $(z-1)$ so that the function becomes nice at $z=1$ ? At first glance, it might seem like you want to do it three times so that it becomes $\log z$ , and this is definitely nice at $z=1$ . But actually, just two times is enough! This is because $\log z$ has a zero at $z=1$ .

How can we verify that $\frac{\log z}{z-1}$ is nice at $z=1$ ? Meaning, does this have a limit when we send $z \to 1$ ? One way to do this is using the Taylor series for $\log(1-z)$ . We can write
$\log z = \log(1-(1-z)) = -(1-z) - \frac{(1-z)^2}{2} - \frac{((1-z)^3}{3} - \ldots,$ so that
$\frac{\log z}{z-1} = \log(1-(1-z)) = 1 + \frac{z-1}{2} + \frac{(z-1)^2}{3} + \ldots.$ This clearly does not explode at $z=1$ , so this proves that $\frac{\log z}{z-1}$ is continuous at $z=1$ .

So $n=2$ is the power we need.

Second step: We now have to use the formula.

$(z-1)^2 \cdot f(z)$ is just $\frac{\log z}{z-1}$ .

Now we differentiate this $2-1 = 1$ times. By the quotient rule, this is equal to $\frac{\frac{z-1}{z} - \log z}{(z-1)^2}.$

Now we need to take the limit of this as $z \to 1$ . Blegh. Let's first make a substitution $w=z-1$ . This will make things a bit simpler to write.
$\lim_{z \to 1} \frac{\frac{z-1}{z} - \log z}{(z-1)^2} = \lim_{w \to 0} \frac{\frac{w}{1+w} - \log(1+w)}{w^2} = \lim_{w \to 0} \frac{w - (1+w)\log(1+w)}{(1+w)w^2}$ We can find this limit using the Taylor expansion for $\log(1+w)$ . The expression is equal to
$\frac{w - (1+w)(w - w^2/2 + \text{blah})}{(1+w)w^2} = \frac{-w^2/2 + \text{blahblah}}{(1+w)w^2} = \frac{-1}{2(1+w)} + \text{blahblahblah}.$ The "blah" terms go to 0 fast enough so that we can ignore them (I'm not being very rigorous). So if we send $w \to 0$ here, we get $\boxed{\operatorname*{Res}_{z=1} \frac{\log z}{(z - 1)^3} = \frac{-1}{2}}$ .

Lesson 7: The Residue Theorem

We're finally here.

The Residue Theorem gives us a very simple way to evaluate basically any contour integral!

To reduce complexity, the theorem will only be for "simple loops", which are basically loops that don't intersect themselves. (So circles are good, but figure-8's are not.)

Theorem (Residue Theorem): Let $\gamma$ be a simple looping path, which loops in the counter-clockwise direction. Let $f$ be a holomorphic (meromorphic) function that is not defined (i.e. has a "pole") at some points $z_1,z_2,\cdots,z_k$ inside the loop. Then
$\int_\gamma f(z)\,dz = 2\pi i \sum_{j=1}^k \operatorname*{Res}_{z=z_j} f(z).$ In other words, the contour integral is just $2\pi i$ times the sum of all residues inside the loop.

Example 1

Let us return to this classic example of integrating $1/z$ around the unit circle.
$\int_{\partial D} \frac{1}{z}\,dz$ There is exactly one point in the interior of the circle where $1/z$ explodes, and that is $z=0$ . The residue of $1/z$ at $z=0$ , as we saw before, is $1$ . Thus we may apply the Residue Theorem to get
$\int_{\partial D} \frac{1}{z}\,dz = 2\pi i \times \operatorname*{Res}_{z=0} \frac1z = \boxed{2\pi i}.$
Example 2

Let us return to Lesson 3 Example 2, where we once again integrate $1/z$ , but this time around a square.

$[asy] size(7cm); draw((-2,0)--(2,0),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-2)--(0,2),p=linewidth(1.5),arrow=Arrows(5)); draw((0,-1)--(1,-1)--(1,1)--(-1,1)--(-1,-1)--(0,-1),p=red,arrow=MidArrow); label("$\Gamma$",(-1,1.5),p=red); [/asy]$

What is $\int_\gamma \frac1z\,dz$ ? Again, there is only one place inside this square where the integrand explodes: $z=0$ . The residue there has not changed: It's still $1$ . So $\int_\gamma \frac1z\,dz = \boxed{2\pi i}$ , just like in the previous example.

This shows that the integral of $1/z$ around any path that loops around $0$ will be $2\pi i$ . (As long as you loop only once, and you loop in the counter-clockwise direction.)

Example 3

Let $\gamma$ be a (simple) loop oriented counter-clockwise. Let $f$ be a holomorphic function that is defined everywhere in the interior of this loop.

Then the sum of the residues inside the loop is... well, $0$ , because there are no singularities to deal with (and hence no residues to sum up). So $\int_\gamma f(z)\,dz = 0$ . This is Cauchy's Theorem.

Example 4

Let $\partial D(r)$ be the circle centered at $0$ of radius $r$ , oriented counter-clockwise. When $r \neq 1$ , what are the possible values of the integral
$\int_{\partial D(r)} \frac{1}{z+1/z}\,dz?$ First let us write it as
$\int_{\partial D(r)} \frac{z}{z^2+1}\,dz = \int_{\partial D(r)} \frac{z}{(z+i)(z-i)}\,dz.$ This tells us that the integrand explodes at only two points: $z=i$ and $z=-i$ .

First case: If $0 < r < 1$ , then the interior of the disk of radius $r$ won't contain either $i$ or $-i$ , meaning that the integrand is defined everywhere inside it. So by Cauchy's Theorem, we would get $\int_{\partial D(r)} \frac{z}{z^2+1}\,dz = 0$ .

Second case: If $r > 1$ , then the interior of the disk of radius $r$ will contain both $i$ and $-i$ . We need to sum up the residues at both of these singularities.

For $z=i$ , we multiply the integrand by $z-i$ to get $\frac{z}{z+i}$ . Plugging in $z = i$ into this will give us $\frac{i}{2i} = \frac{1}{2}$ , so $\operatorname*{Res}_{z=i} \frac{z}{z^2+1} = \frac12$ .

For $z = -i$ , we instead multiply by $z+i$ to get $\frac{z}{z-i}$ . Plugging in $z = -i$ gives $\frac{-i}{-2i} = \frac12$ , so $\operatorname*{Res}_{z=-i} \frac{z}{z^2+1} = \frac12$ .

Therefore, by the Residue Theorem,
$\int_{\partial D(r)} \frac{z}{z^2+1}\,dz = 2\pi i\left(\operatorname*{Res}_{z=i} \frac{z}{z^2+1} + \operatorname*{Res}_{z=-i} \frac{z}{z^2+1}\right) = 2\pi i(1/2+1/2) = 2\pi i.$
The two possible values are $\boxed{0}$ and $\boxed{2\pi i}$ .

Exercises

Compute $\int_{\partial D} \frac{1}{\sin z}\,dz$ . Answer
$2\pi i$
Compute $\int_{\partial D(\pi+0.001)} \frac{1}{\sin z}\,dz$ . (Careful!) Answer
$-2\pi i$
Solution
There are actually three singularities you need to worry about: $-\pi$ , $0$ , and $\pi$ . Since $\lim_{z \to \pi} \frac{z-\pi}{\sin z} = -1$ , we have $\operatorname*{Res}_{z=\pi} \frac{1}{\sin z} = -1$ , and similarly for $-\pi$ . The integral is then $2\pi i(-1 + 1 - 1) = -2\pi i$ .
Compute $\int_\gamma \frac{e^z}{z^4-1}\,dz$ , where $\gamma$ is the rectangle with vertices $\pm 100 \pm 0.5i$ , oriented counter-clockwise.
Solution
The singularities are $z = -1,1,i,-i$ . But only two of them lie inside the interior of the looping rectangle: $-1$ and $1$ .

The residue at $1$ is just $\frac{e^z}{(z+1)(z+i)(z-i)}$ evaluated at $z=1$ , which is $\frac{e^{1}}{2(1+i)(1-i)} = \frac{e}{4}$ .

The residue at $-1$ is just $\frac{e^z}{(z-1)(z+i)(z-i)}$ evaluated at $z=-1$ , which is $\frac{e^{-1}}{-2(-1+i)(-1-i)} = \frac{1}{-4e}$ .

By the Residue Theorem, we conclude that
$\int_\gamma \frac{e^z}{z^4-1}\,dz = \boxed{2\pi i\left(\frac{e}{4} - \frac{1}{4e}\right)}.$

(Next Part)

This post has been edited 2 times. Last edited by greenturtle3141, May 19, 2023, 5:55 AM

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Turtle Math

Complex Numbers II: A Guide to Contour Integration (Part 2 of 3)

by greenturtle3141, May 19, 2023, 5:09 AM

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