Question

\[\int_{\gamma}\frac{1}{4z^2-1}dz\] where \(\gamma\) is the unit circle, \(z \in \mathbb{C}\).

Answer 1

z is the unit circle in which plane?

Answer 2

I said this is equal to
\[\int_{0}^{2\pi}\frac{ie^{it}}{4(e^{it})^2-1}dt\] and let u=e^(it)
Then I get \[\int_{1}^{1}\frac{du}{4u^2-1}dz\] which should equal 0.. but is this possible because there are roots -1/2 and +1/2 inside of the closed countour \(\gamma\) so how can it be 0 if it's not satisfying the criteria for Cauchy's Theorem?

Answer 3

complex plane

Answer 4

No idea... Sorry man I don't know integration with complex numbers, except the simplest case.