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I have been asked to work out $\oint_{|𝑧|=2Ο€}\tan(𝑧) dz$ 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: $2Ο€ 𝑖 [Res(3Ο€/2) + Res(Ο€/2) + Res(βˆ’Ο€/2) + Res(βˆ’3Ο€/2)]$ 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:

$Res(Ο€/2) = \lim_{𝑧→π/2} (𝑧 βˆ’ Ο€/2)\frac{\sin𝑧}{\cos𝑧}$

$= \lim_{𝑧→π/2} (𝑧 βˆ’ Ο€/2) \frac{\sin 𝑧} {\cos 𝑧 βˆ’ \cos(Ο€/2)}$ (introducing \cos(Ο€/2) = 0 judiciously)

$= \lim_{𝑧→π/2}\frac{\sin𝑧}{\frac{\cos π‘§βˆ’\cos(Ο€/2)}{(π‘§βˆ’Ο€/2)}}$

$= \lim_{𝑧→π/2}\frac{\sin𝑧}{(\cos𝑧)β€²}$ (by definition of the derivative)

$= \lim_{𝑧→π/2}\frac{\sin𝑧}{βˆ’\sin𝑧} = βˆ’1$.

However I cant follow the lines:

$= \lim_{𝑧→π/2}\frac{\sin𝑧}{\frac{\cos π‘§βˆ’\cos(Ο€/2)}{(π‘§βˆ’Ο€/2)}}$

$= \lim_{𝑧→π/2}\frac{\sin𝑧}{(\cos𝑧)β€²}$ (by definition of the derivative)

How is $\frac{\cos π‘§βˆ’\cos(Ο€/2)}{(π‘§βˆ’Ο€/2)}=(\cos z)'$? 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 $βˆ’8Ο€i$

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

2 Answers 2

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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 $h=\pi/2-x$

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  • $\begingroup$ 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! $\endgroup$
    – Mackie
    Apr 21, 2017 at 13:36
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    $\begingroup$ In your limit $h$ is wrong set, instead $z$ must tend to $\pi/2$ approximating from its left. And yes! You got it! $\endgroup$
    – HBR
    Apr 21, 2017 at 13:45
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This is showing the general method that the residue of $\frac{N}{D} = \frac{N}{D'}$ if it exists at that point.

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