By the way, the original problem could have been done very simply in the following way:
Let u= z-\pi. Then z= u+ \pi and the function becomes
\frac{sin(z)}{z-\pi}= \frac{sin(u+\pi)}{u}
But sin(u+\pi)= sin(u)cos(\pi)+ cos(u)sin(\pi)= -sin(u). Thus,
\frac{sin(z)}{z- \pi}= -\frac{sin(u)}{u}[/itex]<br />
As z goes to \pi, u goes to 0 and the fraction clearly goes to -1. Since the limit exists that is a removable singularity.<br />
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Oh, and <b>please</b> use parentheses. It was not at all clear to me, at first, whether you meant sin(z)/(z-\pi) or sin(z/(z-\pi)).<br />
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The Laurent series of an analytic function or a function with a removeable singularity does not have any negative powers of z, of course. <br />
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Since, as I said before, <br />
\frac{sin(z)}{z- \pi}= -\frac{sin(u)}{u}[/itex]<br />
You can find the &quot;Laurent&quot; series for sin(z)/(z-\pi) by writing out the Taylor&#039;s series for sin(u), dividing by u, and then replacing u by z-\pi.<br />
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Since sin(u) does not have a &quot;constant&quot; term, sin(u)/u will not have any negative powers of u, sin(z)/(z-\pi) will not have any negative powers of z.
