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The integral in question is

$\int_{_C} \frac{z}{cos(z)}\,dz,$ Where C is path $e^{jt},$ where $\ 0< t < 2\pi$

Since the pole of the function is +$\frac{\pi}{2}$ and -$\frac{\pi}{2}$, both of which are outside the path of integration, does that mean the integral equals to zero?

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By Cauchy's, yes. – anon Jun 21 '12 at 9:14
Of course, the integrand has more poles than that – but the rest of them are even further outside the path of integration, so you're safe from them. – Harald Hanche-Olsen Jun 21 '12 at 9:38

Trying to save one more unanswered question from doom:

The poles of the function are in the points where $$\cos z=0\Longleftrightarrow z=\left(n-\frac{1}{2}\right)\pi\,,\,n\in\Bbb Z$$ Since the path is $$C:=\{z\in\Bbb C\;:\; |z|=1\Longleftrightarrow z=e^{it}\,\,,\,0\leq t\leq 2\pi\}$$ the unit circle centered at the origin, and this path contains in its interior no poles of the function, we get by Cauchy's Integral Theorem that $$\oint_C\frac{z}{\cos z}dz=0$$

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Trying to save one more unupvoted answer from doom. – akkkk Nov 27 '12 at 16:07

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