A question doubt

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=reiθz=r e^{i \theta}

The integral becomes ∫0−2πr2eiθdθ\large{\displaystyle \int^{-2 \pi}_{0} r^2 e^{i \theta} d \theta}

Second method (I am stuck in this)

∣z∣|z| is not analytic at z=0z=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.

#Calculus

Note by Tanishq Varshney
5 years, 1 month ago

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1 vote

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Comments

Since ∣z∣|z| is not analytic, our best bet is simply to parametrize the integral. Write z = −1+e−itz \,=\, -1 + e^{-it} for 0≤t≤2π0 \le t \le 2\pi, and the integral becomes ∫C∣z∣ dz=∫02π∣−1+e−it∣(−ie−it) dt  =  −i∫02π2(1−cos⁡t)e−it dt=−i∫02π2sin⁡12te−it dt  =  −∫02π(e12it−e−12it)e−it dt=−[2ie−12it−23ie−32it]02π=83i \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 - 5 years, 1 month ago

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

Tanishq Varshney - 5 years, 1 month ago

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The integral of 1∣z∣\tfrac{1}{|z|} around CC would diverge, since the integrand would become 12cosec 12t e−it\tfrac12\mathrm{cosec}\,\tfrac12t\,e^{-it}, and the real part of that integral would diverge.

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 - 5 years, 1 month ago
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