Evaluate \(\large{\displaystyle \int_{C} |z| dz}\) where \(C\) is the circle given by \(|z+1|=1\) described in clockwise sense.

First method(which I surely think is wrong)

\(z=r e^{i \theta}\)

The integral becomes \(\large{\displaystyle \int^{-2 \pi}_{0} r^2 e^{i \theta} d \theta}\)

Second method (I am stuck in this)

\(|z|\) is not analytic at \(z=0\) as the cauchy reiman conditions are not satisfied. Do we have to apply keyhole contour here and what will be its answer plz help.

## Comments

Sort by:

TopNewestSince \(|z|\) is not analytic, our best bet is simply to parametrize the integral. Write \(z \,=\, -1 + e^{-it}\) for \(0 \le t \le 2\pi\), and the integral becomes \[ \begin{array}{rcl} \displaystyle \int_C |z|\,dz & = & \displaystyle \int_0^{2\pi} \big|-1 + e^{-it}\big| (-ie^{-it})\,dt \; = \; -i \int_0^{2\pi} \sqrt{2(1-\cos t)} e^{-it}\,dt \\ & = & \displaystyle -i \int_0^{2\pi} 2\sin\tfrac12t e^{-it}\,dt \; = \; -\int_0^{2\pi}\big(e^{\frac12it} - e^{-\frac12it}\big)e^{-it}\,dt \\ & = & \displaystyle -\Big[2ie^{-\frac12it} - \tfrac23ie^{-\frac32it}\Big]_0^{2\pi} \\ & = & \tfrac83i \end{array} \] – Mark Hennings · 9 months, 2 weeks ago

Log in to reply

– Tanishq Varshney · 9 months, 2 weeks ago

Thank you sir for the solution, is there any way to test whether the complex integral converge , e.g if we had \(\frac{1}{|z|}\) then?Log in to reply

Since you can simply translate the complex integral into a real integral of a complex function using the parametrization, you can simply test the real integrand for convergence. – Mark Hennings · 9 months, 2 weeks ago

Log in to reply

@Mark Hennings sir,@Brian Charlesworth sir,@Maggie Miller. – Tanishq Varshney · 9 months, 2 weeks ago

Log in to reply