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Evaluate the integral $$ \int_0^{2\pi}\frac{1}{3-2\cos \theta +\sin\theta}\,\mathrm d\theta. $$


This must be solved by using $d \theta = dz/(iz)$ and transforming $\sin, \cos$ to complex form, but I am stuck after transforming it. It is now $$\frac{1}{(2i+1)z^2 - 6i z + (2i+1)}$$ I don't know how to complete it now to find the singularities and solve by the residue theorem.

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  • $\begingroup$ @Epictetus: Why did the OP use the complex-analysis tag??! $\endgroup$ – mrs Dec 18 '12 at 14:15
  • $\begingroup$ this must be solve by using d theta = (dz/iz) and sin , cos will be in Z form , but i am stuck after transforming it to the complex ! $\endgroup$ – lawati Dec 18 '12 at 14:18
  • $\begingroup$ it is now 1/([2i+1]z^2 - 6i z + [2i+1] ) i dont know how to complete it now to find the singularities and solve by resduie $\endgroup$ – lawati Dec 18 '12 at 14:19
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With complex methods we have that $$I=\int_{0}^{2\pi}\frac{dt}{3-2\cos t+\sin t}=\oint_{C}\frac{1}{3-z-z^{-1}+\frac{z-z^{-1}}{2i}}\frac{dz}{iz}=\oint_{C}\frac{1}{3zi-z^2i-i+\frac{z^2-1}2}dz$$ where $C$ is the unit circle.

Consider $$f(z)=\frac{1}{3zi-z^2i-i+\frac{z^2-1}2}=\frac{2}{z^2(1-2i)+6zi-2i-1}$$ A trivial factorisation gives $$f(z)=\frac{2}{(1-2i)(z+\frac{i}{1-2i})(z+\frac{5i}{1-2i})}$$ You can then compute the residues and use the residue theorem

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  • $\begingroup$ thanks , but how you factorize this one ? $\endgroup$ – lawati Dec 18 '12 at 14:48
  • $\begingroup$ With the discriminant that is negative... $\endgroup$ – Nameless Dec 18 '12 at 14:48
  • $\begingroup$ the sin t has a coefficient of 1/2i right ? $\endgroup$ – lawati Dec 18 '12 at 15:20
  • $\begingroup$ $\sin t=\frac{z-z^{-1}}{2i}$ $\endgroup$ – Nameless Dec 18 '12 at 15:21
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Hint: try using the substitution $u=\tan \left(\frac{\theta}{2}\right)$.

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  • $\begingroup$ this must be solve by using d theta = (dz/iz) and sin , cos will be in Z form , but i am stuck after transforming it to the complex ! $\endgroup$ – lawati Dec 18 '12 at 14:15

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