Question

Exercice 1 We consider the function f(x) = 2 #0 and for r > 0. let S, = {€ C/2 = r} with positive orientation. For 0 < <R, we denote by r the curve consisting of SRUT-R,-€) US, UL, R), where S = {z E C/121 = } with negative orientation. 1. Prove that o = [513)dz = [5(=)dz + [s()de – [ (dz + 1" $(x)dr.

Answer 1

An analytic function whose Laurent series is given by

- Σας (-,

(1)

can be integrated term by term using a closed contour encircling $z_0$,

f (2) d 2	$=$	$sum_(n=-infty)^(infty)a_nint_gamma(z-z_0)^ndz$	(2)
	$=$	$sum_(n=-infty)^(-2)a_nint_gamma(z-z_0)^ndz+a_(-1)int_gamma(dz)/(z-z_0)+sum_(n=0)^(infty)a_nint_gamma(z-z_0)^ndz.$	(3)

The Cauchy integral theorem requires that the first and last terms vanish, so we have

$int_gammaf(z)dz=a_(-1)int_gamma(dz)/(z-z_0),$

(4)

where $a_(-1)$ is the complex residue. Using the contour $z=gamma(t)=e^(it)+z_0$ gives

$int_gamma(dz)/(z-z_0)=int_0^(2pi)(ie^(it)dt)/(e^(it))=2pii,$

(5)

so we have

$int_gammaf(z)dz=2piia_(-1).$

(6)

If the contour encloses multiple poles, then the theorem gives the general result

(7)

where is the set of poles contained inside the contour. This amazing theorem therefore says that the value of a contour integral for any contour in the complex plane depends only on the properties of a few very special points inside the contour.

Contour

Outer semi circle corresponds to S​​​​​​R: |z| = R, and inner semi circle corresponds to S​​​​​​e: |z| = ε, and two horizontal line segments [-R, -ε] and [ε,R]

$f(z) = \frac{e^i^z}{z}$

Only pole of f(z) is z=0, which lies outside the contour.

Hence as per Cauchy residue theorem

on poles which lie inside contour

$ f(z)dz = 211 Resf (2)

Therefore

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