- #1
Nikitin
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[SOLVED] Complex contour integral zero while containing a pole?
##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
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?
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?
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