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

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 $$-2n \pi i a$$

• 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 – Conrad Aug 6 at 21:13

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}

• your answer is amazing. I really hoped that you would write where I have been wrong. – hash man Aug 7 at 8:19
• can you explain why in the first methodology you swap the sign? – hash man Aug 7 at 8:30
• What do you mean by "swap the sign?" – Mark Viola Aug 7 at 13:07
• 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. – hash man Aug 7 at 15:13
• Yes. $(0-a)=-a$. – Mark Viola Aug 7 at 15:14

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.

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