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# 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)

$$z=r e^{i \theta}$$

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.

Note by Tanishq Varshney
8 months, 2 weeks ago

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Since $$|z|$$ 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 $\begin{array}{rcl} \displaystyle \int_C |z|\,dz & = & \displaystyle \int_0^{2\pi} \big|-1 + e^{-it}\big| (-ie^{-it})\,dt \; = \; -i \int_0^{2\pi} \sqrt{2(1-\cos t)} e^{-it}\,dt \\ & = & \displaystyle -i \int_0^{2\pi} 2\sin\tfrac12t e^{-it}\,dt \; = \; -\int_0^{2\pi}\big(e^{\frac12it} - e^{-\frac12it}\big)e^{-it}\,dt \\ & = & \displaystyle -\Big[2ie^{-\frac12it} - \tfrac23ie^{-\frac32it}\Big]_0^{2\pi} \\ & = & \tfrac83i \end{array}$ · 8 months, 2 weeks 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 $$\frac{1}{|z|}$$ then? · 8 months, 2 weeks ago

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The integral of $$\tfrac{1}{|z|}$$ around $$C$$ would diverge, since the integrand would become $$\tfrac12\mathrm{cosec}\,\tfrac12t\,e^{-it}$$, 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. · 8 months, 2 weeks ago

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