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$$\int_0^{2\pi} \frac{d \theta}{5-3 \cos \theta}$$

My attempt:

Let $z=e^{i\theta}$ which gives $d\theta = \frac{dz}{iz}$

Thus,

$$\oint_C \frac{1}{5-3(\frac{z+z^{-1}}{2})}\frac{dz}{iz}$$

$$=\frac{1}{i}\oint_C \frac{dz}{(3z-1)(z-3)}$$

We can ignore the singularity at $z=3$ because it lies outside the unit circle and thus we don't need to account for it when computing the residues.

$$=\frac{1}{i}\oint_C f(z) \, dz = 2\pi [\operatorname{Res}(f(z),\frac{1}{3})]$$

$$\operatorname{Res}(f(z),\frac{1}{3})=\frac{\lim_{z\to1/3} (3z-1)\cdot \frac{1}{(3z-1)(z-3)}}{0!}$$

$$=-\frac{9}{8}$$

Which yields:

$$=\frac{1}{i}\oint_C f(z) \, dz = 2\pi [-\frac{9}{8}]$$

$$=-\frac{9\pi}{4}$$

But how can I have a negative answer for an integration??? Did I make a mistake somewhere??

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  • $\begingroup$ Something strange is going on between your second and third integrals. Check your work there. $\endgroup$ – Alex S Apr 2 '16 at 21:35
  • $\begingroup$ $$\left(5-3\frac{z+z^{-1}}{2}\right)z=5z-\frac32z^2-\frac32=\color{red}{-\frac12}(3z-1)(z-3)$$ $\endgroup$ – Did Apr 2 '16 at 21:36
  • $\begingroup$ $$(5z-\frac{3z^2}{2}-\frac{3}{2})=0$$. To simplify finding the roots, we can multiply both sides by $-2$, which yields: $$(3z^2-10z+3)=0$$. Thus, $$(3z-1)(z-3)=0$$ $\endgroup$ – whatwhatwhat Apr 2 '16 at 21:42
  • $\begingroup$ @Did , what is the difference? $\endgroup$ – whatwhatwhat Apr 2 '16 at 21:42
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    $\begingroup$ $$\oint_C \frac{1}{5-3(\frac{z+z^{-1}}{2})}\frac{dz}{iz}=-2\frac{1}{i}\oint_C \frac{dz}{(3z-1)(z-3)}\ne\frac{1}{i}\oint_C \frac{dz}{(3z-1)(z-3)}$$ $\endgroup$ – Did Apr 2 '16 at 21:51
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It is true that in order to find the roots, you can multiply both sides of $$ 5z-\frac{3z^2} 2 - \frac 3 2 = 0 $$ by $-2$, getting $$ -10z + 3z^2 + 3 = 0, $$ and then using your favorite method of solving quadratic equations, concluding that the roots are $3$ and $1/3$.

But that does not mean that $-10z+3z^2 + 3z$ is equal to $(-1/2)(10z + 3z^2 + 3)$ when the value of $z$ is such that the value of $-10z+3z^2 + 3z$ is not $0$.

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  • $\begingroup$ Ohhhhhhh ok that makes sense. It's like I changed the expression to make things simpler, but it is not the same as the original expression at all points of $z$, so I must change it back. $\endgroup$ – whatwhatwhat Apr 2 '16 at 21:58
  • $\begingroup$ For future reference and for others to see why: wolframalpha.com/input/?i=plot+x%5E2,+2x%5E2,+4x%5E2,+8x%5E2 $\endgroup$ – whatwhatwhat Apr 2 '16 at 22:00

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