Question Number 130369 by gowsalya last updated on 24/Jan/21
Answered by Olaf last updated on 24/Jan/21
![Ω = ∫_(∣z∣=1) xdz Ω = ∫_0 ^(2π) cosθd(cosθ+isinθ) Ω = ∫_0 ^(2π) cosθ(−sinθ+icosθ)dθ Ω = ∫_0 ^(2π) (−(1/2)sin2θ+i((1+cos2θ)/2))dθ Ω = [(1/4)cos2θ+i((θ+(1/2)sin2θ)/2)]_0 ^(2π) Ω = iπ](https://www.tinkutara.com/question/Q130376.png)
$$\Omega\:=\:\int_{\mid{z}\mid=\mathrm{1}} {xdz} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}\theta{d}\left(\mathrm{cos}\theta+{i}\mathrm{sin}\theta\right) \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}\theta\left(−\mathrm{sin}\theta+{i}\mathrm{cos}\theta\right){d}\theta \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}\theta+{i}\frac{\mathrm{1}+\mathrm{cos2}\theta}{\mathrm{2}}\right){d}\theta \\ $$$$\Omega\:=\:\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos2}\theta+{i}\frac{\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$\Omega\:=\:{i}\pi \\ $$
Commented by gowsalya last updated on 25/Jan/21
thank u