# Residue theorem to evaluate integrals

#### nacho

##### Active member

For the first question, I have tried looking at examples and have noted that the bounds have been provided in a manner:

like |z|=1 (as given in part ii)

I am not sure how to get transform the given |z-pi|=pi in such a format, although i suspect it would be something like |z|=2pi ?
Where do I go from here?

for part ii) i don't understand how to treat the z^m term. does this imply that z^m is a series expansion, or is it trying to say m is a positive integer?
How do I solve this equation?

Many thanks in advanced MHB, you are more help than my professors at uni! (and mathhelpforum, )

(as per forum rules, i've cross posted this question elsewhere: Residue theorem)

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
In general if you are given something like $$\displaystyle |z-a|=b \,\,,\, b> 0\,\, , \text{ and } z,a\in \mathbb{C}$$ we are referring to the circle of radius $b$ and the center is at $a$.

To see this more clearly let $$\displaystyle z=x+iy \, , \, a=s+it$$ then we have

$$\displaystyle |z-a|=|x+iy -(s+it)|= \sqrt{(x-s)^2+(y-t)^2}$$

So we have

$$\displaystyle (x-s)^2+(y-t)^2=b^2$$

which describes a circle of center $(s,t)$ and radius $b$.

For the special case $$\displaystyle |z|=1$$ just let $s=t=0 , b=1$

so we have $x^2+y^2=1$ which is the unit circle .

For the other part of the question treat $z^m$ as a polynomial of order $m$ clearly this is an entire function so we need not worry about it . You are integrating the function around a circle of radius $1$ so look how the function behaves in and on this contour is it holomorphic or does it have poles ?

#### nacho

##### Active member
In general if you are given something like $$\displaystyle |z-a|=b \,\,,\, b> 0\,\, , \text{ and } z,a\in \mathbb{C}$$ we are referring to the circle of radius $b$ and the center is at $a$.

To see this more clearly let $$\displaystyle z=x+iy \, , \, a=s+it$$ then we have

$$\displaystyle |z-a|=|x+iy -(s+it)|= \sqrt{(x-s)^2+(y-t)^2}$$

So we have

$$\displaystyle (x-s)^2+(y-t)^2=b^2$$

which describes a circle of center $(s,t)$ and radius $b$.

For the special case $$\displaystyle |z|=1$$ just let $s=t=0 , b=1$

so we have $x^2+y^2=1$ which is the unit circle .

For the other part of the question treat $z^m$ as a polynomial of order $m$ clearly this is an entire function so we need not worry about it . You are integrating the function around a circle of radius $1$ so look how the function behaves in and on this contour is it holomorphic or does it have poles ?

ahh, that makes sense, thank you.

So for part i)
my integral is centred at pi, with a radius of pi.
the function has a pole at pi/2, so I apply the residue formula,
Res(f,c)
=>
Res(f,pi/2) = $\frac{1}{2\pi i}$ ... ?

as for part ii)

i THINK this is a trick question. distant memories of l'hopitals and squeeze theorem are coming to me, but i do believe that sin(1/0) is defined. so this function has no poles?

in that case, what do i do?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
ahh, that makes sense, thank you.

So for part i)
my integral is centred at pi, with a radius of pi.
the function has a pole at pi/2, so I apply the residue formula,
Res(f,c)
=>
Res(f,pi/2) = $\frac{1}{2\pi i}$ ... ?
How exactly did you get that ?

as for part ii)

i THINK this is a trick question. distant memories of l'hopitals and squeeze theorem are coming to me, but i do believe that sin(1/0) is defined. so this function has no poles?

in that case, what do i do?
You need to use the Laurent expansion , the function has an essential singularity at $0$.

#### nacho

##### Active member
How exactly did you get that ?

You need to use the Laurent expansion , the function has an essential singularity at $0$.
Residues are just going completely over my head.

was your query for part i) in regards to what i said about the poles?

The poles are the points where there is a hole in the given domain, no?

as for part ii) I am completely and utterly lost.

Is the laurent expansion used on z^m?
afterwards, won't we get an infinite series (which i guess would be convergent)?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
i ) We are integrating the function $z \tan(z)$ . Clearly the function can be rewritten as $$\displaystyle \frac{z\, \sin(z)}{\cos(z)}$$ and we only care about the poles of $\cos(z)$ in the interval $$\displaystyle [0,2\pi ]$$ , so what are the singularities at this interval ?

#### nacho

##### Active member
OH, over [0,2pi] the singularities are pi/2 and 3pi/2

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
So we need to find the residues at both $$\displaystyle z=\frac{\pi}{2},\frac{3\pi}{2}$$ using the function $$\displaystyle \frac{z \, \sin(z)}{\cos(z)}$$ , how do you do that ?

Remember that the numerator is holomorphic and not-zero at both points.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Consider the following function $$\displaystyle f(z)=\frac{g(z)}{h(z)}$$ and we want to find the residue at $$\displaystyle z=z_0$$ assume that $$\displaystyle g(z_0)\neq 0,h(z_0)=0$$

Assuming that the function $g(z)$ is holomorphic at $z_0$ .

Then we can say the following

$$\displaystyle \text{Res}(f;z_0)=\lim_{z \to z_0} (z-z_0) \frac{g(z)}{h(z)-h(z_0)}= \frac{g(z_0)}{\lim_{z \to z_0}\frac{h(z)-h(z_0)}{z-z_0}}=\frac{g(z_0)}{h'(z_0)}$$

#### nacho

##### Active member
Oh, ok.

I ended up getting -pi/2 and -3pi/2 for my answers.
is this correct?

Could i get some assistance on part ii)

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Have you learnt the Laurent expansion ? Can you expand $\sin\left(\frac{1}{z} \right)$ around $0$?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
That is a different question ! We are looking at the Laurent expansion of $\sin(1/z)$ . It is essential when finding residues to be able to expand using the Laurent expansion because we are only interested in the coefficient of the term $$\displaystyle \frac{1}{z-z_0}$$ .

Since we know the Taylor expansion of $$\displaystyle \sin(z)$$ around $0$

$$\displaystyle \sin(z) = z-\frac{z^3}{3!}+\frac{z^5}{5!}\,+\cdots$$

$$\displaystyle \sin(1/z)= \frac{1}{z}-\frac{1}{3!\, z^3}+\frac{1}{5! z^5}+\cdots$$

$$\displaystyle z^m \sin(1/z)= z^{m-1}-\frac{z^{m-3}}{3!\, }+\frac{z^{m-5}}{5! }+\cdots$$

So what will be the term that contains $$\displaystyle \frac{1}{z}$$ ?

#### nacho

##### Active member
wont it be z^(m-1)

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
wont it be z^(m-1)
This will be true if $m=0$ so we get $$\displaystyle \frac{1}{z}$$ but we might get others remember that we are choosing $m$ is arbitrary number so we need to test all the values .
For example if you take the term $$\displaystyle z^{m-3}$$ if you let $m=2$ we get $1/z$ can you conjecture a general formula ?

#### nacho

##### Active member
This will be true if $m=0$ so we get $$\displaystyle \frac{1}{z}$$ but we might get others remember that we are choosing $m$ is arbitrary number so we need to test all the values .
For example if you take the term $$\displaystyle z^{m-3}$$ if you let $m=2$ we get $1/z$ can you conjecture a general formula ?
I cant't, I'm a little lost.

What did you mean precisely when you asked "What will be the term that contains $\frac{1}{z}$ ?

Do you mean, which of the a0, a1, a2...an it would be?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
The problem here is the term $$\displaystyle z^m$$ which will tell us where the residue is but since $m$ is arbitrary we have to find a general formula , let us take an easy example :

$$\displaystyle \int_{\gamma(0,1)}z^m \, dz \,\,\,, \,\, m \in \mathbb{Z}$$

How to solve this integral ?