Complex contour integral zero while containing a pole?

[Nikitin]

[SOLVED] Complex contour integral zero while containing a pole?

Homework Statement

$f(z) = \frac{1}{z^2 +2z +5} = \frac{1}{(z-z_1)(z-z_2)}$, where $z_1= -1+2i$ and $z_2 = -1-2i$.

Now, let z be parametrized as $z(\theta)=Re^{i \theta}$, where $\theta$ can have values in the interval of $[0,\pi]$. Furthermore, let $R \rightarrow \infty$ and $a>0$. Show $$\lim_{R \to \infty} \int_{S_R} f(z) e^{iaz} dz = 0$$

The Attempt at a Solution

My main problem with this, is that the upper half-plane contains a residue for $f(z) e^{iaz}$, namely at $z_1$, and thus the contour-integral can impossibly be zero...

But on the other hand, I can see from the ML-inequality theorem that the sum of the integral should go towards zero when R goes towards infinity.. Help?

 

[Nikitin]

edit nevermind, I just noticed that the parametrization is not a closed curve but rather a semi-circle. forget this thread.