The question was to evaluate the integral of f(z) dz, around C, where C is the unit circle centered at the origin, using the general cauchy's theorem.
f(z) is 1/((2z^2)+1).
I know that f(z) is not analytic at i(sqrt2)/2 and -i(sqrt2)/2 and both points happen to be inside C, so cauchy's theorem can't be applied.
What's my next step?
I don't see how you can use Cauchy's theorem here... unless maybe you mean Cauchy's residue theorem.
But that's not necessary either. It's probably easiest to use Cauchy's integral formula. Draw a line through the unit circle such that each pole is inside one of the resulting half-circles. Letbe appropriately-oriented paths corresponding to each half-circle such that
. Then by Cauchy's integral formulae we have
![=2\pi i\left[\frac{1}{2(-[i/\sqrt{2}]-[i/\sqrt{2}])}+\frac{1}{2([i/\sqrt{2}]+[i/\sqrt{2}])}\right]](http://latex.codecogs.com/png.latex?=2\pi i\left[\frac{1}{2(-[i/\sqrt{2}]-[i/\sqrt{2}])}+\frac{1}{2([i/\sqrt{2}]+[i/\sqrt{2}])}\right])
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