Evaluate $\oint_{|z|=1} \frac{dz}{\sqrt{6z^2-5z+1}} $ I want to evaluate $\oint_{|z|=1} \frac{dz}{\sqrt{6z^2-5z+1}} $ (Joseph Bak's chapter 12 example 2)
According to the textbook, I need to change the contour to $R$ instead of $1$. But I couldn't find reason why $\oint_{|z|=1} \frac{dz}{\sqrt{6z^2-5z+1}} = \oint_{|z|=R} \frac{dz}{\sqrt{6z^2-5z+1}}$ holds for $R$ greater than 1. Can you please give me reason why it holds?
 A: Answer has been edited:
It's a common trick in complex analysis to parameterise the contour, and take the limit as the contour approaches some other shape (which in general will change the value of the integral, but under certain conditions it will not change the value of the integral). For example, it's used a lot when using complex integration methods to evaluate real valued integrals.
Now, in this case, we are integrating the function ${\frac{1}{\sqrt{6z^2-5z+1}}}$. This has two singularities which lie in the unit circle, and you can use Residue Theorem to evaluate this.
What the authour has done - he has decided to replace the unit circle with a circle of radius ${R>1}$. This is valid since, as I said before, our function is analytic on the region ${\{z \in \mathbb{C}\ |\ 1\leq |z| < R\}}$. This is indeed a Theorem (Deformation Theorem, give that bad boy a Google ;D). Since ${R>1}$ can be arbitrarily large, he's approximated the function on the $R$ circle to just ${\frac{1}{\sqrt{6}z}}$ (I guess an intuitive way of thinking about it is that as ${R}$ get's larger and larger, the function becomes closer to being ${\frac{1}{\sqrt{6}z}}$ on the ${R}$ circle). This isn't rigorous, although I'm sure it could probably be rigoursly justified.
Anyway, the integral of ${\oint_{|z|=R}\frac{1}{z}dz=2\pi i}$, and hence he's gotten
$${\oint_{|z|=1}\frac{1}{\sqrt{6z^2-5z+1}}dz=\oint_{|z|=R}\frac{1}{\sqrt{6z^2-5z+1}}dz\rightarrow\frac{1}{\sqrt{6}}\oint_{|z|=R}\frac{1}{z}dz=\frac{2\pi i}{\sqrt{6}}}$$
Where the ${\rightarrow}$ is being used to mean ${R\rightarrow \infty}$
