# Are the Poles inside the contour?

I'm trying to evaluate this integral using the Residue theorem (It doesn't look to hard) $$\int_{0}^{2\pi}\frac{ dx}{(a+b\cos(x))^2}$$ (where $a>b>0$ ) I do the standard substitution of $z=e^x$ and get $dx=-idz/z$ and $\cos(x)=\frac{z+z^{-1}}{2}$, put it in my integral and, with a little algebra, get

$$-4i \oint_{|z|=1} \frac{zdz}{(bz^2+2az+b)^2} \,$$

But when I try find the poles I end up getting:

$$z_{1,2}=\frac{-2a\pm \sqrt{4a^2-4b^2}}{2b} \overset{\mathrm{c=a/b}}{=}-c\pm \sqrt{c^2-1}$$

but now I can't find which pole is inside the contour and which isn't becouse:

for $z_1$

$-c+\sqrt{c^2-1}<1$

$\sqrt{c^2-1}<1+c$

$c^2-1<1+2c+c^2$

$-2<2c$

and since our $c>1$ this pole is inside the contour. On the other hand for $z_2$ we have

$-c-\sqrt{c^2-1}<1$

$-\sqrt{c^2-1}<c+1\implies \sqrt{c^2-1}>-c-1$

since $c>1>0$ this is equal to saying $-5<6$ or $-6<5$ it will always be true! Is that a problem (is it a sign of a mistake) or is the pole at $z_2$ also inside the contour?

• $-c-\sqrt{c^2-1}<-1$, so it's outside – user8268 Jan 31 '18 at 19:07
• so my condition should actually be $|-c\pm \sqrt{c^2-1}|<1$ ? – Alexandar Solženjicin Jan 31 '18 at 19:29
• Have a look at page 46 of my notes for alternative approaches. I like to regard such integral as a multiple of the area enclosed by some ellipse. – Jack D'Aurizio Jan 31 '18 at 19:39
• Wow, thank you very much :) – Alexandar Solženjicin Jan 31 '18 at 19:47

$z_2\not\in B(0,1)$. Indeed, if $z_2\in B(0,1)$ then $$|z_2|<1$$ $$c+\sqrt{c^2-1}<1$$ $$1+\sqrt{c^2-1}<c+\sqrt{c^2-1}<1$$ $$\sqrt{c^2-1}<0$$ which is a contradiction.
• as $c=a/b$ and $a>b>0$ entonces $c>1$ – Mauricio Ruiz Feb 1 '18 at 15:43
• I agree, but how does $1+\sqrt{c^2-1}<c+\sqrt{c^2-1}<1$ imply that if $1+\sqrt{c^2-1}<1$ then $c+\sqrt{c^2-1}<1$ ? – Alexandar Solženjicin Feb 1 '18 at 15:46
• we have $$c+\sqrt{c^2-1}<1$$. As $c=a/b$ then $c>1$, then $$1+\sqrt{c^2-1}<c+\sqrt{c^2-1}$$. By transitivity $$1+\sqrt{c^2-1}<1$$. So $$\sqrt{c^2-1}<0$$ – Mauricio Ruiz Feb 1 '18 at 15:53