Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2014

Consider a circle $|Z|=1$ which has center $(0,0)$ & radius $=1$

Put $z=re^{i\theta}=1e^{i\theta}=e^{i\theta}\\ \therefore dz=e^{i\theta}.id\theta=iz\space d\theta\\ \therefore d\theta=\dfrac {dz}{iz}$

Also $,\cos \theta =\dfrac {e^{i\theta}+e^{-i\theta}}2=\dfrac {z+z^{-1}}2=\dfrac {z^2+1}{2z}$

And $,\cos2\theta=R.P(e^{i2\theta})=R.P.(e^{i\theta})^2=R.P.(z^2)$

On substituting we get,

$$\int\limits_0^{2\pi}\dfrac {\cos2\theta}{5+4\cos\theta}.d\theta$$

$=\int\dfrac {R.P.(z^2)}{5+4[\frac {(z^2+1)}{2z}]}.\dfrac {dz}{iz}\\ =R.P\int \dfrac {z.z^2}{5z+2z^2+2}.\dfrac {dz}{iz}\\ = R.P.\int\dfrac {z^2.dz}{i(2z^2+4z+z+2)}\\ =R.P.\int \dfrac {z^2.dz}{i(2z+1)(z+2)}$

Hence $z_0=\dfrac {-1}2$ and $z_0=-2$ are simple poles

$z_0=-2$ lies outside while $z_0=\dfrac {-1}2$ lies inside the circle $|z|=1$

$R_1=$ Residue of $f(z) at\space \space z_0=\dfrac {-1}2\\ =\lim\limits_{z\to z_0}(z-z_0)\times f(z)\\ =\lim\limits_{z\to \frac {-1}2}[z+\dfrac 12]\times \dfrac {z^2}{i\times 2(z+\frac 12)(z+2)}\\ =\lim\limits_{z\to \frac {-1}2}\dfrac {z^2}{2i(z+2)}\\ =\dfrac {(\frac {-1}2)^2}{2i(\frac {-1}2+2)}\\ =\dfrac 1{12i}$

By Cauchy’s residue theorem

$$\int f(z).dz=2\pi i(R_1+R_2)$$

$\therefore R.P. \int \dfrac {z^2.dz}{i(2z+1)(z+2)}\\ =R.P.[2\pi i\times \dfrac 1{12i}]\\ \therefore \int\limits_0^{2\pi}\dfrac {\cos2\theta}{5+4\cos \theta}.d\theta\\ =R.P. (\dfrac {\pi6})\\ =\dfrac \pi6\\ \therefore \int\limits_0^{2\pi}\dfrac {\cos2\theta}{5+4\cos \theta}.d\theta\\ =\dfrac \pi6$