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$\int\limits_{S(1,1)} (\frac{z}{z-1} )^n dz$ where $S(c,r)$ refers to a 2-dimensional sphere with a radius(r) and a center(c). My thoughts till now are $\left(\frac{z}{z-1} \right)^n=(z^n)\left(\frac{1}{z-1} \right)^n$ and write $\frac{1}{z-1} = \sum c_{k}z^n$ as a series with center $0$ so I can use the common $z$ in the above product $(z^n)\left(\frac{1}{z-1} \right)^n=z^n\left(\sum c_{k}z^k\right)^n=\sum \left(c_{k}\right)^nz^{nk+n}$. What I got was the series $\left(\sum_k z^{k+1}\right)^n$. And then I tried integrate that. Am I taking it too far, cause I think this may be simpler.

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You could also apply Cauchy's integral formula, assuming $n \in \mathbb{N}\setminus \{0\}$ $$f^{(n)}(a)=\frac{n!}{2\pi i} \int\limits_{S(a,r)}\frac{f(z)}{(z-a)^{n+1}}dz$$ In this case, the function is $f(z)=z^n$ $$\int\limits_{S(1,1)}\frac{z^n}{(z-1)^n}dz=\int\limits_{S(1,1)}\frac{f(z)}{(z-1)^n}dz=\frac{2\pi i}{(n-1)!}f^{(n-1)}(1)=2\pi i \frac{n!}{(n-1)!}=2\pi i \cdot n$$

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I presume you mean the path integral $$\oint_{S(1,1)} \left(\frac{z}{z-1}\right)^n\; dz$$ where $S(c,r)$ is the circle $|z - c| = r$ in the positive (counterclockwise) direction, and $n$ is a positive integer.

It's better to use a series in powers of $z-1$. If you take $w = z-1$, you get $$ \int_{S(1,1)}\left(\frac{z}{z-1}\right)^n\; dz = \int_{S(1,0)}\left(1 + \frac{1}{w}\right)^n \; dw= \sum_{j=0}^n {n \choose j} \int_{S(1,0)} w^{-j}\; dw = \ldots$$

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  • $\begingroup$ So I can assume that since the only anomaly-point in $S(0,1)$ is its center $0$ and only when $j=1$. All the integrals with $j \neq 1$ are nulled since $w^{-j}$ defines a holomorphic function except $j=1$ where we have the anomaly. So we just have to calculate $(\binom n 1 )*2πι$? $\endgroup$ – Pookaros Jan 14 '18 at 18:40
  • $\begingroup$ $w=0$ is still a singularity of $w^{-j}$ for $j \ne 1$, but the integral is $0$ because for $j \ne 1$, $w^{-j}$ has an antiderivative $w^{1-j}/(1-j)$ that is holomorphic except at $w=0$. $\endgroup$ – Robert Israel Jan 14 '18 at 18:54

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