More about residue calculation

Ok, what about the case of, say, tan(z) = \frac{sin(z)}{cos(z)} at latex z = \frac{\pi}{2}} $? One way: note lim_{z \rightarrow \frac{\pi}{2}} \frac{sin(z)(z-\frac{\pi}{2})}{cos(z)} can be evaluated by L’Hopital’s rule which leads to lim_{z \rightarrow \frac{\pi}{2}}\frac{sin(z)+(z-\frac{\pi}{2})cos(z)}{-sin(z)} =-1 and note how the product rule lead to the numerator being what it was. In general, if we are interested in \frac{h(z)}{g(z)} where h(z_0) \neq 0, g(z_0) = 0 The residue will be given by \frac{h(z_0)}{g'(z_0)} provided g'(z_0) \neq 0 ; that is, provided g has a zero of order 1 at z_0

Many residue calculation theorems work in such a manner.

In general, if f has a pole of order m at z=z_0 then f(z) = \frac{b_m}{(z-z_0)^m} + \frac{b_{m-1}}{(z-z_0)^{m-1}} + ...+\frac{b_1}{z-z_0} + .... so if we multiply both sides by (z-z_)^m and then differentiate repeatedly and focus on the residue: b_1(z-z_0)^{m-1} gets differentiated m-1 times to yield (m-1)(m-2)....(2)(1) b_1 so b_1 = \frac{1}{(n-1)!}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^{m-1}f(z)

Example: \frac{1}{z^2 (z+1)} has a pole of order 2 at the origin. So the residue: (\frac{z^2}{z^2(z+1)})' = -\frac{1}{(z+1)^2} which is -1 at z = 0 . That is simpler than calculating the series.

There are many other formulas that we could develop but most are some combination of the ideas that we just discussed. And sometimes, it is best to just grit your teeth and calculate the series.


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