Author Topic: 2.3 Q6  (Read 87 times)

syee

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2.3 Q6
« on: November 08, 2018, 01:19:53 AM »
I don't know where to go with the integrand after I've converted sin and cos into 1/2i(z-1/z) and 1/2(z+1/z) respectively.

Amy Zhao

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Re: 2.3 Q6
« Reply #1 on: November 08, 2018, 02:46:03 AM »
$\int_0^{2\pi} \frac{dθ}{3+sinθ+cosθ}~dx$$=\int_{|z|=1} \frac{dz}{iz(3+\frac{1}{2i}(z-\frac{1}{z}))+\frac{1}{2}(z+\frac{1}{z}))}$
where $iz(3+\frac{1}{2i}(z-\frac{1}{z})+\frac{1}{2}(z+\frac{1}{z})) = i(3z+(\frac{z^2}{2i}-\frac{1}{2i})+(\frac{z^2}{2}+\frac{1}{2}))$
$=i(\frac{6zi+z^2-1+z^2i+i}{2i})$
$=\frac{(i+1)z^2+6iz+(i-1)}{2}$
The integral $=\int_{|z|=1} \frac{dz}{\frac{(i+1)z^2+6iz+(i-1)}{2}}$
$=\int_{|z|=1} \frac{2dz}{(i+1)z^2+6iz+(i-1)}$
$=\int_{|z|=1} \frac{2dz/(z-(\frac{-3i+i\sqrt{7}}{1+i}))}{z-(\frac{-3i-i\sqrt{7}}{1+i})}$
$So,f(\frac{-3i-i\sqrt{7}}{1+i})=\frac{2}{\frac{-2i\sqrt{7}}{1+i}}$
The integral$=\frac{1+i}{-\sqrt{7}}2\pi$