Q:

Complex integral (Cauchy's Theorem?)

I have $$\int_{\gamma}\frac{1}{4z^2-1}dz$$, where $\gamma$ is the unit circle in the complex plane. 
I said this integral equals to $$\int_{0}^{2\pi}\frac{ie^{it}}{4(e^{it})^2-1}dt$$
Then I let $u=e^{it}$
which gave me
$$\int_{1}^{1}\frac{du}{4u^2-1}$$
 which should equal 0.. but is this possible because there are roots -1/2 and +1/2 inside of the closed contour $\gamma$ so how can it be 0 if it's not satisfying the criteria for Cauchy's Theorem?

A:

It might be easier to expand the integrand as
$\frac{1}{4z^2-1} =  {1 \over 4} \left( \frac{1}{z-{1 \over 2} }-\frac{1}{z+{1 \over 2} } \right)$. (Hence the integral is zero.)

