Solving Simple Poles: Why Does $\frac{e^{ikz}}{z^{2}+m^{2}}$ Have One?

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nklohit
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In the mathematical text I've read, it says that [tex]\frac{e^{ikz}}{z^{2}+m^{2}}[/tex] has only one simple pole, that is, [tex]im[/tex], if [tex]k>0[/tex]. Why?
Has it got 2 simple poles, [tex]I am[/tex] and [tex]-im[/tex] ?
 
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You are correct. It has simple poles at [tex]im[/tex] and [tex]-im[/tex]. Either your text is wrong or perhaps there's a restriction on the domain that you didn't notice.
 
It does have two simple poles at I am and -im, but if k is positive the contour will be closed upwards (the contour's edges run along with x-axis, then form a half circle in the upper plane; taking the radius of the half circle to infinity, we find that this contour is exactly equal to the integral we are interested in) and the only pole enclosed will be the one at +im.

Therefore, there is only one pole that contributes to the residue.