Mastering Residue Calculation Techniques in Complex Variables

Greetings students and welcome back to another video on complex variables. In this lesson, I'm going to state and

prove three well-known techniques that allow you to find the residues of a complex function. This is going to come

in handy when applying the residue theorem and when integrating complicated real functions as we'll see in the next

video. So, let's start. Suppose I have a complex function with the following la. Recall from my Laurance series video

that a Luron series is essentially a generalized tailaylor series that includes rational expressions which have

negative powers on Z. In other words, the B series or the principal part in addition to the polomial portion that

you find in a typical tailaylor series or the analytical part, this a series. Recall also that the residue of a

function f of z at a particular point z knot is the coefficient b1 in the lauron expansion of f of z around z knot. So

this B1 right here, the coefficient of the 1 / Z minus Z knot term. This very brief explanation actually gives rise to

the first technique used to find the residue of f of Z at Z knot, which just involves finding the Lauron expansion of

f of Z around Z knot and then just taking out the B1 from that lauron expansion. I don't really need to prove

this considering that this is after all the definition of the residue. So let's just skip straight to an example. In

this example, we want to find the residue of sin z / z ^2 at z= 0. This is fairly simple. We know that the

tailaylor expansion or the power series representation for sin z about z= 0 is given by z - z cub over 3 factorial plus

z 5 factorial and so on. That means the luron expansion of sin z / z ^ 2 would just be this tailaylor expansion of sin

z / z ^ 2. And if we simplify everything then we'll get 1 / z minus z over 3 factorial plus z cubed over 5 factorial

and so on and the powers will keep getting higher as we keep going. So the residue of sin z over z ^2 at 0 is just

the coefficient of the 1 / z minus 0 or 1 / z term which in this case is just one.

The second technique I'm going to talk about doesn't involve finding the lauron expansion. It applies to cases where you

can't find the Lauron expansion or you just don't want to. It specifically applies to when your function f of z has

a simple pole at z knot and you want to find the residue at that simple pole. Recall that by a simple pole I mean a

pole about which the lauron expansion of f of z only goes up to the 1 / z minus z kn term in the lauron expansion. In

other words, this summation in the principal part up here only has b1 as the non-trivial term. Everything else is

just zero. The second technique is also pretty simple. All you do is multiply the function f of z by z minus z knot

and take the limit as z approaches z knot. Proving this technique is also fairly simple. So let's suppose for the

proof that f of z has a lauron expansion about z knot given by the following expression. You know the one we

mentioned above. Now because z knot is a simple pole of f of z. This b series here only goes up till j = 1. So the

laur expansion of f of z can be simplified to the following. The sum from k= 0 to infinity of a k * z - z k

plus b1 / z - z kn. So now the principal part contains just the b1 term. Let's now multiply both sides by z minus

z kn to get the following equation. Now what happens when we take the limit of both sides as Z approaches Z knot? Well,

this first summation term goes away since it contains powers of Z minus Z KN greater than or equal to 1. On the other

hand, the second term, the constant B1, the residue, it just stays there. So we can say that the residue of a complex

function f of z at a simple pole z is just the limit as z approaches z knot of zus z kn * f of z. Now just to clarify

the reason we use the limit is that f of z is actually undefined at z knot. Z knot is still a pole after all even

though it's a simple pole and if you multiply f of z by z minus z knot and evaluate the whole result at z knot you

would just end up with the 0 / 0 which is undetermined. So that's why we've used the limit. So you technically don't

get 0 / 0 and everything turns out okay. Let's now do an example to illustrate this technique. In this example, we want

to find the residue of cossine z / z 4 - 1 at z = i. Let's begin by factoring the denominator. So when we multiply by z -

z kn, which in this case is zus i, then we can more easily cancel out the terms. You know that z 4 - 1 is just z ^ 2 - 1

* z ^2 + 1 and that z ^2 + 1 is just z - i * z + i. So let's just plug this factored form into the equation for f of

z and this is what we'll get. Now because cossine z is continuous and differentiable or holorphic at z= i.

That means z= i must be a simple pole since z minus i only appears once in the denominator.

So that means we can use technique number two to find the residue at z= i because z= i is a simple pole and

technique number two only applies to finding residues at simple poles. In this case, the residue is just the

limit as Z approaches I of Z - I * cosine Z / Z ^ 2 - 1 * Z - I Z + I. And if we cancel the Z minus I terms, we end

up with this equation. Now, if we want to take the limit of this expression at Z= I, we can do that very easily just by

substituting in Z= I. The denominator becomes -2 * 2 I, which is -4 I. But the tricky bit is with the numerator. And

even if you try to use a calculator that has complex numbers, you end up with a math error. Still, there's a trick you

can use to evaluate cosine I using the oiler formula. And with the oiler formula, you can isolate and find an

expression for cosine z in terms of just exponentials. I leave it up to the viewer to show that cossine z is e i z +

e i z / 2. It's kind of like cossine x which is e to x + e x / 2. You might be able to see the analogy there. If you

use this equation for cosine z, then you'll find that cossine i is just e + e inverse / 2. So the residue of cosine z

over z 4 - 1 at z= i is just e + e inverse over 8 i. And that solves our example. Now, before using technique

number two, you might wonder, how do I know if the function has a simple pole at Z= Z KN? What if it's a pole of

higher order? Or what if it doesn't even have poles? How can I find that out before using technique number two? Well,

the answer lies in the Lauron series* ZUS Z KN expression I wrote up here, which I'll copy paste below. Now if I

didn't have a pole at Z knot, then all my B coefficients would be zero since my function would be completely analytic at

that point. Therefore, if I then took the limit of this expression as Z approaches Z KN, then I would just end

up with a zero residue. However, if my Z knot was a higher order pole, then in addition to the B1 term in

my B series, I would also have a B2 term and even terms beyond B2 up to the order of the pole. Of course, note that the

terms beyond B2 have one less power on the Z minus Z knot in the denominator, mainly because we've already multiplied

it out once. Now if I take the limit of this expression as Z approaches Z knot then

because of these extra terms corresponding to the higher order pole we get infinity for the limit. Therefore

if I then took the limit of Z minus Z knot * f of z when z knot is a higher order pole I get infinity. So that's how

you can tell you have a simple pole. If this limit is zero then the function doesn't have a pole at z knot and it's

totally analytic there. When this limit is infinity, the pole at Z knot is a higher order pole. And when the limit is

a nonzero finite number, then you do have a simple pole at Znot. And the limit is just the residue B1. So this

quantity, this limit isn't just a way to calculate residues for simple poles. It's also a diagnostic tool to check

whether or not you even have a simple pole at Znot. And that completes the discussion for

technique number two. Let's go to the third and final residue calculation technique. And the second technique you

saw just applied to simple pole zen knot. This technique number three is an extension and it now applies to higher

order poles. It's not as easy as the other ones. And let me begin first by describing how it works.

Suppose you want to find the residue of the function f of z at z= z knot. And z knot turns out to be a non-simple pole.

Turns out to be a higher order pole. In fact, suppose that the order of the pole z is n. So what you do is you start by

multiplying f of z by z minus z kn to the power m where m is some integer that's greater than or equal to the

order of the pole n. Then what you do is you differentiate zus z to the m * f of z m minus1 times and the third step is

to divide that derivative by m -1 factorial and then evaluate the result at the pole z knot and then this final

quantity is your residue at the nons simple pole z knot. Now when it comes to proving this technique the book I was

reading just said that the proof was very simple and left it as an exercise to the reader. I'm not like that. Well,

sometimes. So, following the famous words of the tank commander/ headcoach legend Byron Scott, I'm going to man up

and prove this technique to you. Let's start with the lauron series of f of z. Now, when znot is a pole of order n, we

can expand out the b series so that it goes until the b subn term. Now, let's start with the first step of

technique three. multiply f of z by z minus z kn the power m where m is an integer greater than or equal to n. Now

because m is greater than or equal to n all these powers of z minus z knot in the principal series the b series all

those powers are going to be positive which makes differentiation more straightforward. Now let's apply the

second step and differentiate this whole expression m minus1 times. If we do that then all the powers on the z minus z kn

terms in this a series will reduce by m minus one and we'll also be multiplying the coefficient a subk by k + m * k + m

-1 * all the way to k + 2. This is just using the chain rule and applying it to the differentiation of polomials.

But what about the principal part? Well, if you differentiate the principal part mus1 times, then every term which has a

power less than mus1 will become zero. That's because if you differentiate a polomial, the degrees or powers on that

polomial are going to keep decreasing with every differentiation until you get a constant term and when you

differentiate that constant term, you get zero. Therefore, all the terms in the B series except for B1 become zero.

The B1 term on the other hand is going to have its power reduced by the number of times we differentiate. So, m minus

one. In addition, because the power comes down as a coefficient every time we differentiate, we'll also have mus1

factorial multiplying b1. And now it's time for the final step of the technique where we evaluate the derivative at z= z

knot and divide by mus1 factorial. If we evaluate the derivative at z= z knot, every term in this a series is going to

be zero, leaving only the b1 term. and dividing that by m minus one factorial will just give us the residue b1 at z= z

knot. So that's it. We've proven the validity of technique number three that if you have a higher order pole at z

knot, you can multiply that function with that higher order pole whose residue you want to find by z minus z

kned to the power of an integer m greater than or equal to the order of the pole. Then you can differentiate

that m minus one times and then divide by m minus one factorial and evaluate the derivative at that higher order pole

z knot to get the residue. Let's finish off the video with one last example applying this third technique.

In this example, we want to find the residue of z cosine z / zus p<unk> cubed at z= pi. We can already see from the

function that the numerator z cosine z doesn't have any poles at z= pi which means that the denominator is the only

factor contributing to the z= pi pole and since the denominator contains zus p<unk> cubed we can conclude that z=

p<unk> is a third order pole. So let's use technique number three and multiply f of z by zus p<unk> cubed leaving us

with only z cosine z. Now because we multiplied by Z -<unk> cubed, we have to differentiate this expression 3 -1

times. In other words, we find the second derivative. You can use the product rule to show that the second

derivative here is just -2 sin zus z cosine z. Finally, in the third step, we'll substitute z=<unk> and divide by

the factorial of 3 - 1, which is just 2. Sin pi is zero while cosine pi is 1. So for the residue, we'll end up with

p<unk> by 2. And that should do it for the video. If you enjoyed the lecture, feel free to like and subscribe. In the

next complex variables video, we'll talk about how to use the residue theorem to determine integrals, a video that some

of you have already requested. Thank you for watching and this is the faculty of Khan signing out.