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calculate $\oint_{|z|=1}\left(\frac{z}{z-a}\right)^{n}$ whereas a is different from 1, and n is integer. My try:

\begin{align} & \oint_{|z|=1}\left(\frac{z}{z-a}\right)^n \, dz\\[6pt] & \oint_{|z|=1}\frac{z^n \, dz}{\sum_{k=0}^n z^k(-a)^{n-k}}\\[6pt] & \oint_{|z|=1}\sum_{k=0}^n\frac{z^n \, dz}{z^k(-a)^{n-k}}\\[6pt] & \oint_{|z|=1}\sum_{k=0}^n\frac{z^{n-k} \, dz}{(-a)^{n-k}} \end{align}

I forgot here to write the binomial but it won't change the answer.

no singularities therefore it's $0.$ if $n$ is smaller than $0$: \begin{align} & \oint_{|z|=1}\left(\frac{z-a}{z}\right)^n \, dz\\[6pt] & \oint_{|z|=1}\frac{\sum_{k=0}^n \binom n k z^k(-a)^{n-k} \, dz}{z^n}\\[6pt] & \oint_{|z|=1}\sum_{k=0}^n \binom n k \frac{z^{k}(-a)^{n-k}dz}{z^{n}}\\[6pt] & \oint_{|z|=1}\sum_{k=0}^n \binom n k z^{k-n}(-a)^{n-k}\, dz \end{align} so if so we look $k-n=-1$ $k=n-1$ which is exactly $-na$ therefore the value of the integral is $-2n \pi i a$

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    $\begingroup$ one can write $(\frac{z}{z-a})^{n}=(1+a/(z-a))^n=\frac{na}{z-a}+g(z)$ where $g$ has an antiderivative since is a Laurent polynomial with no degree $-1$ term; so for example if $|a| >1$, the function is analytic in the unit disc hence the integral is zero etc $\endgroup$ – Conrad Aug 6 at 21:13
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If $|a|>1$, and $n$ is a non-negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ is analytic on the disk $|z|\le 1$ and Cauchy Integral Theorem guarantees that

$$\oint_{|z|=1}\frac{z^n}{(z-a)^n}\,dz=0$$


METHODOLOGY $1$: For $|a|<1$

For $n\ge 0$, we can write

$$\left(\frac{z}{z-a}\right)^n=\left(1+\frac{a}{z-a}\right)^n=\sum_{k=0}^n \binom{n}{k}a^k(z-a)^{-k}$$

The coefficient on the $\frac{1}{z-a}$ term is $na$. So from the Residue Theorem, we have for $|a|<1$

$$\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz=2\pi i na$$


For $n<0$, we can write

$$\left(\frac{z}{z-a}\right)^n=\left(1+\frac{a}{z}\right)^{|n|}=\sum_{k=0}^{|n|} \binom{|n|}{k}(-a)^k(z)^{-k}$$

The coefficient on the $\frac{1}{z}$ term is $-|n|a$. So from the Residue Theorem, we have

$$\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz=-2\pi i |n|a=2\pi i n a$$



METHODOLOGY $2$:For $|a|<1$

If $n$ is a negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ has a pole of order $|n|$ at $z=0$. Hence, application of the residue theorem reveals

$$\begin{align} \oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz&=\oint_{|z|=1}\left(\frac{z-a}{z}\right)^{|n|}\,dz\\\\ &=2\pi i \frac1{(|n|-1)!}\lim_{z\to 0}\frac{d^{|n|-1}}{dz^{|n|-1}}(z-a)^{|n|}\\\\ &=-2\pi i|n|a\\\\ &=2\pi i na \end{align}$$


If $|a|<1$, and $n$ is a non-negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ has a pole of order $n$ at $z=a$ and application of the residue theorem reveals

$$\begin{align} \oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz&=2\pi i \frac1{(n-1)!}\lim_{z\to a}\frac{d^{n-1}}{dz^{n-1}}z^{|n|}\\\\ &=2\pi i n a \end{align}$$

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  • $\begingroup$ your answer is amazing. I really hoped that you would write where I have been wrong. $\endgroup$ – hash man Aug 7 at 8:19
  • $\begingroup$ can you explain why in the first methodology you swap the sign? $\endgroup$ – hash man Aug 7 at 8:30
  • $\begingroup$ What do you mean by "swap the sign?" $\endgroup$ – Mark Viola Aug 7 at 13:07
  • $\begingroup$ you had $\begin{align} \oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz&=\oint_{|z|=1}\left(\frac{z-a}{z}\right)^{|n|}\,dz\\\\ &=2\pi i \frac1{(|n|-1)!}\lim_{z\to 0}\frac{d^{|n|-1}}{dz^{|n|-1}}(z-a)^{|n|}\\\\ &=-2\pi i|n|a\\\\ &=2\pi i na \end{align}$ and in the 2nd row from the end you took the negative. $\endgroup$ – hash man Aug 7 at 15:13
  • $\begingroup$ Yes. $(0-a)=-a$. $\endgroup$ – Mark Viola Aug 7 at 15:14
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You can't just pull the sum out of the denominator. You need Cauchy's generalized integral formula for $n\geq0$. If $a$ is in the interior of $C$, where $C$ is a circle, then

$$f^{(k)}(a)=\frac{k!}{2\pi\mathrm i}\oint_C\frac{f(z)}{(z-a)^{k+1}}\mathrm dz.$$

Here, $k=n-1$ and $f(z)=z^n$. So the integral is $\frac{2\pi\mathrm i}{(n-1)!}n!a=2\pi\mathrm i na$. This is for $n\geq0$. For $n<0$ what you did looks good (you're not pulling a sum out of the denominator there). But you should be using $-n$ or $\vert n\vert$ when flipping denominator and numerator. So the solution is $2\pi\mathrm i n a$ again.

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  • $\begingroup$ can you explain where is my mistake regarding that I got the opposite sign? $\endgroup$ – hash man Aug 7 at 8:22
  • $\begingroup$ You flipped denominator and numerator without flipping the sign of the exponent. $\endgroup$ – Vercassivelaunos Aug 7 at 9:00

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