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Calculate $$\oint_\gamma \frac 1{z-\sin z} dz$$ where the contour is the unit circle in the complex plane.

I do not know how to find the order of the pole at 0, though I believe it is 3. Once I have such an order, however, I do not even know how to proceed.

Pedro
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Johnny Apple
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1 Answers1

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$$\frac{1}{z-\sin{z}} = \frac{6}{z^3}\frac{1}{1-z^2/20+O(z^4)} = \frac{6}{z^3} \left (1+\frac{z^2}{20}+ O[z^4]\right ) = \frac{6}{z^3} + \frac{3}{10} \frac{1}{z} + O(z)$$

Thus, the residue at $z=0$ is $3/10$, and the integral is equal to $i 3 \pi/5$.

EDIT

$z=0$ is the only pole of the integrand in the unit circle.

Ron Gordon
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  • Where does your first equality come from? – Johnny Apple Aug 29 '13 at 01:37
  • Taylor series for sine: $$\sin{z} = z - \frac16 z^3 + \frac{1}{120} z^5 - \cdots$$ – Ron Gordon Aug 29 '13 at 01:39
  • How did you invert it? What procedure did you use? – Johnny Apple Aug 29 '13 at 01:46
  • Another Taylor series: $$\frac{1}{1-w} = 1+w + O(w^2)$$ – Ron Gordon Aug 29 '13 at 01:48
  • How do you know that $z = 0$ is the only one ( pole ) inside $\left\vert z\right\vert = 1$ ? – Felix Marin Aug 29 '13 at 06:31
  • @FelixMarin: Consider $$\sin{z}-z=i (\cos (x) \sinh (y)-y)+\sin (x) \cosh (y)-x$$ The imaginary part is zero when $\cos (x)=y/\sinh{y}$. Plug this value of $x$ into the real part to get the function $$f(y)=\sin{(\arccos{(y/\sinh{y})})} \cosh{y} - \arccos{(y/\sinh{y})}$$ Plot this function over $y \in (0,1)$ to reveal a zero only at $y=0$. Thus, $z=0$ is the only zero in the unit disk. – Ron Gordon Aug 29 '13 at 07:30