written 5.0 years ago by | • modified 4.9 years ago |
Solution:
Given $I =\int log$ z dz -----(1)
Put z= $e^{i\theta}, r = 1$
diff. w.r.t .$\theta$
$dz =i e^{i\theta} \theta$
since it is unit circle , $\theta $ varies from 0 to 2$\pi $
put z and dz values in equation (1)
$I = \int_0^{2\pi} (log e^{i\theta}) i e^{i\theta} d\theta$
=$ i \int_0^{2\pi} (log e^{i\theta} e^{i\theta} )d\theta$
=$i \int_0^{2\pi} i\theta e^{i\theta} d\theta$
=$i.i \int_0^{2\pi} \theta e^{i\theta} d\theta$
=$-1 \int_0^{2\pi} \theta e^{i\theta} d\theta$
=[$-\theta \frac{e^{i\theta}}{i} - \frac{e^{i\theta}}{i.i}]_0^{2\pi}$
=[-$\theta \frac{e^{i\theta}}{i} - \frac{e^{i\theta}}{-1}]_0^{2\pi}$
= [-$\theta \frac{e^{i\theta}}{i} + {e^{i\theta}}]_0^{2\pi}$
= $[-(\frac {e^{2\pi i}}{i}.2\pi+e^{2\pi i})-(0 \times\frac{e^{i0}}{i} +e^{i \times0})] $
=[-$\frac {2\pi}{i}+1-1)]$ since $e^{2\pi i} =1 $
=$-\frac {2\pi}{i}$
=$-2\pi \times -i $
=$2\pi i $