# Evaluate by means of the Cauchy Residue Theorem.

I have been asked to work out $\oint_{|π§|=2Ο}\tan(π§) dz$ by using the cauchy residue theorem (where the contour is positively oriented), I got that there are singularities at $-3\pi/2, -\pi/2, \pi/2$ and $3\pi/2$. I know that this integral will be equal to: $2Ο π [Res(3Ο/2) + Res(Ο/2) + Res(βΟ/2) + Res(β3Ο/2)]$ by the cauchy residue theorem. However then I am stuck at working out these residues, I looked at the solutions given to this and I got lost these are what is given:

$Res(Ο/2) = \lim_{π§βΟ/2} (π§ β Ο/2)\frac{\sinπ§}{\cosπ§}$

$= \lim_{π§βΟ/2} (π§ β Ο/2) \frac{\sin π§} {\cos π§ β \cos(Ο/2)}$ (introducing \cos(Ο/2) = 0 judiciously)

$= \lim_{π§βΟ/2}\frac{\sinπ§}{\frac{\cos π§β\cos(Ο/2)}{(π§βΟ/2)}}$

$= \lim_{π§βΟ/2}\frac{\sinπ§}{(\cosπ§)β²}$ (by definition of the derivative)

$= \lim_{π§βΟ/2}\frac{\sinπ§}{β\sinπ§} = β1$.

However I cant follow the lines:

$= \lim_{π§βΟ/2}\frac{\sinπ§}{\frac{\cos π§β\cos(Ο/2)}{(π§βΟ/2)}}$

$= \lim_{π§βΟ/2}\frac{\sinπ§}{(\cosπ§)β²}$ (by definition of the derivative)

How is $\frac{\cos π§β\cos(Ο/2)}{(π§βΟ/2)}=(\cos z)'$? The solution also only gave how to work out $\pi/2$ so i'm assuming this method would be equivalent for all the other residues and then just adding them together would give the answer as $β8Οi$

• This is the L'Hospital method.
– user65203
Apr 21, 2017 at 13:24

Any derivative of a function e.g. of one variable can be computed as follows: $$\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$$ In your case $h=\pi/2-x$
• Oh! and because $cosz$ is an even function when we get $f(z)'=lim_{h->0}\frac{f(\pi/2)-f(z)}{\pi/2-z}$ we can just switch over the signs on the top to get this result! Apr 21, 2017 at 13:36
• In your limit $h$ is wrong set, instead $z$ must tend to $\pi/2$ approximating from its left. And yes! You got it!
This is showing the general method that the residue of $\frac{N}{D} = \frac{N}{D'}$ if it exists at that point.