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Question Number 130370 by gowsalya last updated on 24/Jan/21

Answered by TheSupreme last updated on 24/Jan/21

$$z=e^{i\theta }$$$$\mid dz\mid =\mid {ie}^{i\theta }d\theta \mid =d\theta$$$${\int }_0^{2\pi }\mid e^{i\theta }-1\mid d\theta$$$${\int }_0^{2\pi }\sqrt{{(cos\left(\theta \right)-1)}^2+{sin}^2\theta }d\theta$$$${\int }_0^{2\pi }\sqrt{2-2cos\left(\theta \right)}d\theta$$$$\bullet cos\left(2\gamma \right)=1-2{sin}^2\left(\gamma \right)$$$$2{sin}^2\left(\gamma \right)=1-cos\left(2\gamma \right)$$$${\int }_0^{2\pi }2sin\left(\frac{\theta }{2}\right)d\theta ={[-4cos\left(\frac{\theta }{2}\right)]}_0^{2\pi }=8$$$$$$

Commented by gowsalya last updated on 25/Jan/21

Thank u

Answered by Olaf last updated on 24/Jan/21

$$\mathrm{\Omega }={\int }_{\mid z\mid =1}\mid z-1\mid \mid dz\mid$$$$\mathrm{\Omega }={\int }_0^{2\pi }\mid e^{i\theta }-1\mid \mid {ie}^{i\theta }d\theta \mid$$$$\mathrm{\Omega }={\int }_0^{2\pi }\mid 2{ie}^{i\frac{\theta }{2}}\mid \mid \frac{e^{i\frac{\theta }{2}}-e^{-i\frac{\theta }{2}}}{2i}\mid d\theta$$$$\mathrm{\Omega }=2{\int }_0^{2\pi }\mid \sin \frac{\theta }{2}\mid d\theta$$$$\mathrm{\Omega }=2{\int }_0^{2\pi }\sin \frac{\theta }{2}d\theta$$$$\mathrm{\Omega }=4{[-\cos \frac{\theta }{2}]}_0^{2\pi }$$$$\mathrm{\Omega }=8$$

Commented by gowsalya last updated on 25/Jan/21

Thank u

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