We now revisit residue calculus.

Let f(z) be analytic on and inside a closed contour C, except for a finite number of isolated singularities: z= a1, a2, etc.
which are enclosed by C.

Then:

**ò**_{C}^{ }f(z) dz = 2 pi å^{n}_{k = 1}Res f (a**)**_{k}
We now want to elaborate this a bit more by reference to the
diagram shown. In this case we consider the function f(z) is analytic inside
and on the simple
closed
curve C except at a
finite number of specified points: a, b, c, etc. at which there exist residues: a

**, b**_{- 1}**, c**_{- 1}**, etc.**_{- 1}**ò**

_{C}

^{ }f(z) dz = 2 pi [a

**+ b**

_{- 1}**+ c**

_{- 1}**+ …………………….]**

_{- 1}
That is, 2 pi times the sum of the residues at all the
singularities enclosed by C. To ensure this, one would respectively construct
circles C1, C2, C3 etc. as I have done with respective centers at a, b, c etc.
If we take care to do this properly then we can write:

Where:

**ò**_{C}^{ }f(z) dz = ò_{C1}^{ }f(z) dz + ò_{C2}^{ }f(z) dz +**ò**_{C3}^{ }f(z) dz + ..........Where:

ò

_{C1}^{ }f(z) dz = 2 pi a_{- 1}**ò**

_{C2}

^{ }f(z) dz = 2 pi b

_{- 1}**ò**

_{C3}

^{ }f(z) dz = 2 pi c

_{- 1}
So that:

**ò**

_{C}

^{ }f(z) dz= 2 pi [a

**+ b**

_{- 1}**+ c**

_{- 1}**+ ..] = 2 pi (sum of residues)**

_{- 1}*Example 1:*

Evaluate the integral: ò

_{C }^{ }cot (z) dz
f(z) = cot (z)

For which: ò

_{C}^{ }f(z) dz = 2 pi c_{- 1}
Re-write: f(z) = cot (z) = 1/ tan z

For which singularities occur at tan
z = 0

Or: o,

__+__p,__+__2p,__+__3p etc.
Then Res f(z) = 1/ sec

^{2}z**÷**_{ z = + }_{n }**= 1/ (1/ cos**_{p}^{2}z)
= cos

^{2}z**÷**_{ z = + }_{n }**= cos**_{p}^{2}(np)
and: cos

^{2}(np) = 1 at z = (2n + 1) p)/ 2
Therefore: c

**= 1, and**_{- 1}
ò

_{C}^{ }cot (z) dz = 2 pi (1) = 2 pi*Practice Problem*:

Integrate:

∫

_{-}_{¥}^{¥}^{ }x dx**/**(x**- 2x + 2)**^{2}
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