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contour integral around ∞

pantboio

Member
Nov 20, 2012
45
Consider the function
$$f(z)=Log(\frac{z-a}{z-b})$$
where $a,b\in D(0,r)$ , the disc of radius $r$ centered at the origin, open, and $r>0$. Show that $f$ is holomophic in an open subset of complex plane, containing the complement of the disc:$\mathbb C-D(0,r)$ and compute the integral
$$ \oint_{|z|=r}z^nf(z)dz$$

where $n\geq 0$ is an integer.

I'm quite confused about this problem. Because, the $Log$ in the RHS should be the principal branch, i.e. $Log(z)=log|z|+iArg(z)$, with $Arg(z)\in (-\pi,\pi)$ and $Arg(0)=0$. Then i should prove that $\frac{z-a}{z-b}$ does not intersect the real negative axis, but how to do this? However, suppose i could prove the holomorphicity, then i'm left to compute the integral, for which i tought to use the residue formula applied to $\infty$ point, so calling $I$ the integral,
$$I=2\pi i Ind(\gamma,\infty)Res(f,\infty)$$
where $\gamma$ is the contour $|z|=1$, so that $ind(\gamma,\infty)=-1$. But then the computation of residue is very hard for me. Some help?
 
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pantboio

Member
Nov 20, 2012
45
Re: contour integral around $\infty$

maybe i solved the first part, noticing that
$$Arg(\frac{z-a}{z-b})=Arg(z-a)-Arg(z-b)$$
and the RHS is the angle between $z-a$ and $z-b$, so points such that $\frac{z-a}{z-b}$ lie on the negative real axis are those for which holds

$Arg(\frac{z-a}{z-b})=\pi$

which is the segment joining $a$ with $b$. Hence, taking $z$ with $|z|>Max(|a|,|b|)$ i'm sure that $\frac{z-a}{z-b}$ is not a negative real number
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: contour integral around $\infty$

Consider the function
$$f(z)=Log(\frac{z-a}{z-b})$$
where $a,b\in D(0,r)$ , the disc of radius $r$ centered at the origin, open, and $r>0$. Show that $f$ is holomophic in an open subset of complex plane, containing the complement of the disc:$\mathbb C-D(0,r)$ and compute the integral
$$ \oint_{|z|=r}z^nf(z)dz$$

where $n\geq 0$ is an integer.

I'm quite confused about this problem. Because, the $Log$ in the RHS should be the principal branch, i.e. $Log(z)=log|z|+iArg(z)$, with $Arg(z)\in (-\pi,\pi)$ and $Arg(0)=0$. Then i should prove that $\frac{z-a}{z-b}$ does not intersect the real negative axis, but how to do this? However, suppose i could prove the holomorphicity, then i'm left to compute the integral, for which i tought to use the residue formula applied to $\infty$ point, so calling $I$ the integral,
$$I=2\pi i Ind(\gamma,\infty)Res(f,\infty)$$
where $\gamma$ is the contour $|z|=1$, so that $ind(\gamma,\infty)=-1$. But then the computation of residue is very hard for me. Some help?
Also this post is very interesting!... and also hiding some traps!... proceeding 'step by step' first we remember the McLaurin expansion...

$\displaystyle \ln (z-a) = \ln (-a) - \sum_{n=1}^{\infty} \frac{z^{n}}{n\ a^{n}}$ (1)

... and the series (1) converges for |z|<|a|. Because is...

$\displaystyle f(z)= \ln (z-a)- \ln (z-b)$ (2)

... we obtain that f(z) is holomorphic for $|z|< \text {min}\ (|a|,|b|)$ and, almost immediate consequence, for $r< \text {min}\ (|a|,|b|)$, is...

$\displaystyle \oint_{|z|=|r|} z^{n}\ f(z) = 0$ (3)

What happens for $r> \text {min}\ (|a|,|b|)$ will be examined in next posts...


Kind regards


$\chi$ $\sigma$
 

pantboio

Member
Nov 20, 2012
45
Re: contour integral around $\infty$

Also this post is very interesting!... and also hiding some traps!... proceeding 'step by step' first we remember the McLaurin expansion...

$\displaystyle \ln (z-a) = \ln (-a) - \sum_{n=1}^{\infty} \frac{z^{n}}{n\ a^{n}}$ (1)

... and the series (1) converges for |z|<|a|. Because is...

$\displaystyle f(z)= \ln (z-a)- \ln (z-b)$ (2)

... we obtain that f(z) is holomorphic for $|z|< \text {min}\ (|a|,|b|)$ and, almost immediate consequence, for $r< \text {min}\ (|a|,|b|)$, is...

$\displaystyle \oint_{|z|=|r|} z^{n}\ f(z) = 0$ (3)

What happens for $r> \text {min}\ (|a|,|b|)$ will be examined in next posts...


Kind regards


$\chi$ $\sigma$
maybe i have a proof. I simply factorize $z$ instead of $a$ in $z-a$. So: $Log(z-a)=Log(z(1-\frac{a}{z}))=Log(z)+Log(1-\frac{a}{z})$, and similarly for $Log(z-b)$. Summarizing we have

$$Log(\frac{z-a}{z-b})=Log(z)+Log(1-\frac{a}{z})-Log(z)-Log(1-\frac{b}{z})$$

which holds for $|z|>a,b$, as we wanted,and so..

$$\ldots=\sum_{n=0}^{\infty}\frac{a^n}{z^n} -\sum_{n=0}^{\infty}\frac{b^n}{z^n}$$

Hence we get

$$z^nLog(\frac{z-a}{z-b})=z^n (-2-\frac{a-b}{z}-\frac{a^2-b^2}{z^2}\ldots)$$

we are interested in the coefficient of $\frac{1}{z}$, which is $-(a^{n+1}-b^{n+1})$, but with the signum $+$, since we are looking for residue at $\infty$, not at $0$.

Conclusion: the integral is $2\pi i$ times $a^{n+1}-b^{n+1}$ times the winding number of $\infty$ with respect to the contour, which is counterclockwise oriented, so -1.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: contour integral around $\infty$

maybe i have a proof. I simply factorize $z$ instead of $a$ in $z-a$. So: $Log(z-a)=Log(z(1-\frac{a}{z}))=Log(z)+Log(1-\frac{a}{z})$, and similarly for $Log(z-b)$. Summarizing we have

$$Log(\frac{z-a}{z-b})=Log(z)+Log(1-\frac{a}{z})-Log(z)-Log(1-\frac{b}{z})$$

which holds for $|z|>a,b$, as we wanted,and so..

$$\ldots=\sum_{n={\color{red}1}}^{\infty}\frac{a^n}{{\color{red}n}z^n} -\sum_{n={\color{red}1}}^{\infty}\frac{b^n}{{\color{red}n}z^n}$$

Hence we get

$$z^nLog(\frac{z-a}{z-b})=z^n (-\frac{a-b}{z}-\frac{a^2-b^2}{{\color{red}2}z^2}\ldots)$$

we are interested in the coefficient of $\frac{1}{z}$, which is $\color{red}{-\dfrac{a^{n+1}-b^{n+1}}{n+1}}$, but with the signum $+$, since we are looking for residue at $\infty$, not at $0$.
I think that is correct, except that you have left out the coefficient $1/n$ in the $n$th term of the power series (see the bits in red above). Also, the power series starts with the term $n=1$, not $n=0$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: contour integral around $\infty$

maybe i have a proof. I simply factorize $z$ instead of $a$ in $z-a$. So: $Log(z-a)=Log(z(1-\frac{a}{z}))=Log(z)+Log(1-\frac{a}{z})$, and similarly for $Log(z-b)$. Summarizing we have

$$Log(\frac{z-a}{z-b})=Log(z)+Log(1-\frac{a}{z})-Log(z)-Log(1-\frac{b}{z})$$

which holds for $|z|>a,b$, as we wanted,and so..

$$\ldots=\sum_{n=0}^{\infty}\frac{a^n}{z^n} -\sum_{n=0}^{\infty}\frac{b^n}{z^n}$$

Hence we get

$$z^nLog(\frac{z-a}{z-b})=z^n (-2-\frac{a-b}{z}-\frac{a^2-b^2}{z^2}\ldots)$$

we are interested in the coefficient of $\frac{1}{z}$, which is $-(a^{n+1}-b^{n+1})$, but with the signum $+$, since we are looking for residue at $\infty$, not at $0$.

Conclusion: the integral is $2\pi i$ times $a^{n+1}-b^{n+1}$ times the winding number of $\infty$ with respect to the contour, which is counterclockwise oriented, so -1.
With the exception of some minor details that have been adjusted, Your solution is very good!... we have...

$\displaystyle f(z) = \ln \frac{z-a}{z-b} = \ln (1-\frac{a}{z}) - \ln (1- \frac{b}{z}) = - \sum_{k=1}^{\infty} \frac{a^{k}}{k\ z^{k}} + \sum_{k=1}^{\infty} \frac{b^{k}}{k\ z^{k}}$ (1)

What is important to say is that f(z) is the sum of two Laurent expansions and the convergence condition is $\displaystyle |z|> \text{max} (|a|,|b|)$, and if it is satisfied we easily obtain that...

$\displaystyle \oint_{|z|=r} z^{n}\ f(z)\ dz = 2\ \pi\ i\ \frac {b^{n+1} - a^{n+1}}{n+1}$ (2)

Very well!... we have computed the integral for $\displaystyle r< \text{min} (|a|,|b|)$ and $\displaystyle r> \text{max} (|a|,|b|)$... but what does if happen in other cases?...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: contour integral around $\infty$

Very well!... we have computed the integral for $\displaystyle r< \text{min} (|a|,|b|)$ and $\displaystyle r> \text{max} (|a|,|b|)$... but what does if happen in other cases?...
If we suppose that |b|>|a| [if not we swap b and a of course...], then the only controbution to the integral is given by...


$\displaystyle f_{a} (z) = \ln (z-a) = \ln z + \ln (1-\frac{a}{z}) = \ln z - \sum_{k=1}^{\infty} \frac{a^{n}}{n\ z^{n}}$ (1)

The controbution of the second term of (1) has been already computed, so that we have to find the integral...


$\displaystyle \oint_{|z|=r} z^{n}\ \ln z\ dz$ (2)


... where |a|< r < |b|. Setting $\displaystyle z= r\ e^{i\ \theta}$ we have...


$\displaystyle \oint_{|z|=r} z^{n}\ \ln z\ dz = i\ r^{n+1}\ \int_{- \pi}^{+ \pi} (\ln r + i\ \theta)\ e^{i\ (n+1)\ \theta}\ d \theta = 2\ \pi\ i\ r^{n+1}\ \frac{(-1)^{n+1}}{n+1}$ (3)

... so that for |a|< r < |b| is...

$\displaystyle \oint_{|z|=r} z^{n}\ f_{a}(z)\ dz = 2\ \pi\ i\ \frac{(-1)^{n+1}\ r^{n+1} - a^{n+1}}{n+1} $ (4)


Kind regards

$\chi$ $\sigma$
 
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