I need (just) a hint for finding residue by either using Laurent series or by using limit formula for poles of kth order. I do realize that $z_0 = 0$ is a simple pole, but limit formula is not quite elegant when using it.
Finding residue for $z_0 = 0$ in $f(z) = {{\sin z - z}\over {z^3( z^2 + 9z)}}$
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complex-analysis
laurent-series
1 Answers
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One might recall that
$$\sin z=z-\frac16z^3+\frac1{120}z^5-\dots$$
Thus, you should be able to see that
$$\frac{\sin z-z}{z^4(z+9)}=\frac1{z+9}\left(-\frac16\frac1z+\frac1{120}z-\dots\right)$$
which is pole of order one at $z=0$.
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0As far as I know, term ${1}\over {z + 9}$ is not considered a part of Laurent series at $z = 0$ so I cant find residue at that point. Are you implying that I should find that on my own? Btw, thanks for a hint! – 2017-01-11
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0@GreatDuke Mhm. If you want the residue, all you have to do is multiply by $z$ and take the limit as you usually would, and ofc the $\frac1{z+9}$ causes no problems (but its still part of the residue). – 2017-01-11
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0Nice, I hadn't thought about implementing both formulas at the same time. I figured it out. – 2017-01-11
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0Great then! Happy to help :-P @GreatDuke – 2017-01-11