0
1.4kviews
Show that $\int_c log z dz= 2 \pi i $ where c is the unit circle in z-plane.
1 Answer
0
65views

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 $

Please log in to add an answer.