Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: Dec 2015

Statement:- If F(z) is analytic inside and on a closed curve C of a simply connected region R and if z${}_{0}$ is any point within C, then

Proof :-Since f(z) is analytic inside and on C, $\frac{f(z)}{2-z_0}$ is also analytic inside and on C except at z=z${}_{0}$

We draw a small curve C and C$_{1}$

Hence by Cauchy's extended theorem

$\int \frac{f(z)}{2-z_{0} } dz=\int _{c_{1} }\frac{f(z)}{2-z_{0} } dz $

Putting 2-z$_{0}=re^{i\theta}$ and dz=rie${}^{i\theta}$ d$\theta$ we get on C$_{1}$

$\int _{c_{1} }\frac{f\left(z\right)}{2-z_{0} } dz=\int \frac{6(z_{0} +re^{i\theta })}{re^{i\theta } } \cdot rie^{i\theta }d\theta \\ =\int _{c_{1} }f(z_{0} +re^{i\theta }) id\theta$

As r$ \rightarrow $0 the circle tends to points z${}_{0}$

$\therefore \int _{c_{1} }f\left(z_{0} +re^{i\theta } \right)i\theta \\ =\int _{c_{1} }f(z_{0})id\theta \\ =if (z_{0})\int _{c_{1} }d\theta \\ =i f (z_{0})\int _{0}^{2\pi }d\theta =2\pi i f(z_{0} ) $

Hence we get $\int _{c}\frac{f\left(z\right)}{2-z_{0} } dz =2 \pi i f \left(z_{0} \right)$

Denominator is z${}^{2}$-4

equating it with 0 we get z=$\pm 2$

Now the circle |z-2|=5

It has centre at (2,0) and radius 5

$\therefore $ z=$\pm 2$ both lie inside the circle.

Let z=2, we get 8=4B

$\therefore$ B=2

Let z=-2, we get 4=-4A

$\therefore$ A=-1

$\therefore I=\int _{c}\frac{-1}{z+2} +\frac{2}{z-2} dz$

By Cauchy integral formulas.

$I=-1 2\pi i+2(2\pi i)\\ \therefore I=2\pi i$