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

Marks: 5M

Year: May 2015

Comparing it with polar coordinate $|z| = 1$ i.e. r = 1 $|z| = r$

$z = re^{i\theta}$

Now diff. z w.r.t. $\theta$ we get,

$\frac{dz}{d\theta} = re^{i\theta} \times i \\ \therefore dz = rie^{i\theta} d\theta = ie^{i\theta}dz$ where r = 1

From the given eq. we get

$\therefore$ The given Eq. $\int\limits_{c}|z| dz$ is replaced by $\int\limits_{\frac{3\pi}{2}} ^{\frac{\pi}{2}} 1ie^{i\theta}d\theta$

$\therefore$ Limits is because only left half of unit circle is required

$\therefore i\bigg[\frac{e^{i\theta}}{i}^{\frac{\pi}{2}}\bigg] = e^{\frac{\pi}{2}} - e^{\frac{3\pi i}{2}}$

$\frac{3\pi}{2} \\ = cos\frac{\pi}{2} + isin \frac{\pi}{2} - cos\frac{3\pi}{2} - isin\frac{3\pi}{2} \\ = 0 + i - 0 - (-i) \\ = zi$