Use the substitution $z = e^{i\theta}$ to evaluate
$$\int_{0}^{2\pi} \frac{d\theta}{\sin(\theta)-2}$$
Can somebody point me in the right direction?
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Pointing: $\mathrm{e}^{\mathrm{i}\theta} = \cos \theta + \mathrm{i} \sin \theta$. – Eric Towers Dec 17 '15 at 18:17
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Combine the highlights of Mary Star's and Dr. MV's answers.
Substitute $z=e^{i\theta}$: $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}\theta}{\sin(\theta)-2} &=\oint\overbrace{\frac{2i}{z-\frac1z-4i}}^{\frac1{\sin(\theta)-2}}\overbrace{\ \ \ \ \frac{\mathrm{d}z}{iz}\ \ \ \ }^{\mathrm{d}\theta}\\ &=\oint\frac{2\,\mathrm{d}z}{z^2-4iz-1}\\ &=\oint\frac1{i\sqrt3}\left(\color{#C00000}{\frac1{z-i\left(2+\sqrt3\right)}}-\color{#00A000}{\frac1{z-i\left(2-\sqrt3\right)}}\right)\mathrm{d}z\\ &=\frac{2\pi i}{i\sqrt3}(\color{#C00000}{0}-\color{#00A000}{1})\\[3pt] &=-\frac{2\pi}{\sqrt3} \end{align} $$ where $i\left(2+\sqrt3\right)$ is outside the unit circle and $\frac1{z-i\left(2-\sqrt3\right)}$ has a residue of $1$.
robjohn
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the residue of $f$ at $z=\alpha$ is the coefficient of $\frac1{z-\alpha}$ in the Laurent expansion of $f$ at $\alpha$. – robjohn Dec 18 '15 at 01:09
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Let $z=e^{i\theta}$ so that
$$\begin{align} \int_0^{2\pi} \frac{1}{\sin \theta -2}\,d\theta &=\oint_{|z|=1}\frac{2}{\left(z-i(2+\sqrt 3)\right)\left(z-i(2-\sqrt 3)\right)}\,dz \tag 1\\\\ &=2\pi i \left(\frac{2}{-i2\sqrt 3)}\right) \tag 2\\\\ &=-\frac{2\pi}{\sqrt 3} \end{align}$$
where in going from $(1)$ to $(2)$, we invoked the Residue Theorem.
Mark Viola
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1You seem to be missing a minus sign (the integrand is negative everywhere) – robjohn Dec 17 '15 at 19:09
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@robjon Rob. Thanks for the catch. +1 when one is careless and overlooks the easy things while doind lots of arithmetic in one's head, ... well you get the point. Edited. - Mark – Mark Viola Dec 17 '15 at 20:19
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1@yotengounlcd For $z=e^{i\theta}$, $|z|=1$ and as $\theta$ goes from $0$ to $2\pi$, the entire unit circle is traced. – Mark Viola Dec 17 '15 at 20:21
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Hint:
$$e^{i\theta}=\cos \theta +i \sin \theta$$ $$e^{-i\theta}=\cos \theta -i \sin \theta$$
So $$\sin \theta=\frac{e^{i\theta}-e^{-i \theta}}{2i}\Rightarrow \sin \theta=i\frac{e^{-i\theta}-e^{i \theta}}{2}=i\frac{z^{-1}-z}{2}$$
Mary Star
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