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Use Cauchy's integral formula to evaluate the following integral, $$\int \limits_{\Gamma} \frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-1)(z-2)}dz$$where the contour $\Gamma$ is parameterised by $\gamma : [-\pi,\pi] \rightarrow \mathbb{C}$ given by $\gamma (\theta)=3e^{i\theta}+1$.

If you make $$f(z)=\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-2)}$$ it wont be holomorphic so how can you do it? When z=2, it would be undefined. and 2 is in the region of gamma.

Also correct me if I am wrong but the region is just a circle centred at 1 and radius 3.

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  • $\begingroup$ What does the Cauchy integral formula tell us and how does that help here? $\endgroup$ – dustin Mar 8 '15 at 18:15
  • $\begingroup$ Either split the contour into two circles or use partial fractions. $\endgroup$ – anon Mar 8 '15 at 18:17
  • $\begingroup$ $$f(w)=\frac1{2\pi i} \int \limits_{\Gamma} \frac{f(z)}{z-w}dz$$ where f(z) is holomorphic and w is an interior point. If you make f(z) which I said and w=1, it wont be holomorphic. $\endgroup$ – snowman Mar 8 '15 at 18:17
  • $\begingroup$ Note that the integrand is holomorphic except for simple poles at $1$ and $2$, and these poles both lie in the interior of the region bounded by $\Gamma$. Does this ring a bell? $\endgroup$ – MPW Mar 8 '15 at 18:18
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Note

$$\frac{1}{(z - 1)(z - 2)} = \frac{1}{z - 2} - \frac{1}{z - 1},$$

so if $f(z) = \sin(\pi z^2) + \cos(\pi z^2)$, you can write

$$\int_{\Gamma} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{(z - 1)(z - 2)}\, dz = \int_{\Gamma} \frac{f(z)}{z - 2} \, dz - \int_{\Gamma} \frac{f(z)}{z - 1}\, dz.$$

Since $f$ is entire and both $1$ and $2$ lie inside $\Gamma$, the Cauchy integral formula gives

$$\int_{\Gamma} \frac{f(z)}{z - 1} = 2\pi i f(1) = -2\pi i$$

and

$$\int_{\Gamma} \frac{f(z)}{z - 2} = 2\pi i f(2) = 2\pi i.$$

Hence

$$\int_{\Gamma} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{(z - 1)(z - 2)}\, dz = (2\pi i) - (-2\pi i) = 4\pi i.$$

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  • $\begingroup$ what is your f(z)? $\endgroup$ – snowman Mar 8 '15 at 18:43
  • $\begingroup$ @snowman as I wrote, $f(z) = \sin(\pi z^2) + \cos(\pi z^2)$. $\endgroup$ – kobe Mar 8 '15 at 18:44
  • $\begingroup$ where does the half come from? when you don't have the 1/2, it equates to the LHS... (first line) $\endgroup$ – snowman Mar 8 '15 at 18:50
  • $\begingroup$ @snowman that was an error on my part. Thanks, I made the corrections. $\endgroup$ – kobe Mar 8 '15 at 18:52

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