# 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.

Advertisements