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Question Number 113091 by bobhans last updated on 11/Sep/20

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{curves}$$$$\arg \left(\mathrm{z}\right)=\frac{\pi }{3};\arg \left(\mathrm{z}\right)=\frac{2\pi }{3}\mathrm{and}\arg (\mathrm{z}-2-2\mathrm{i}\sqrt{3})=\pi$$$$\mathrm{on}\mathrm{the}\mathrm{complex}\mathrm{plane}?$$

Answered by john santu last updated on 11/Sep/20

$$\left(i\right)arg\left(z\right)=\frac{\pi }{3}\to \frac{y}{x}=\tan \left(\frac{\pi }{3}\right)$$$$y=x\sqrt{3}$$$$\left(ii\right)arg\left(z\right)=\frac{2\pi }{3}\to \frac{y}{x}=\tan \left(\frac{2\pi }{3}\right)$$$$y=-x\sqrt{3}$$$$\left(iii\right)arg(z-2-2i\sqrt{3})=\pi$$$$\frac{y-2\sqrt{3}}{x-2}=\tan \pi \to y=2\sqrt{3}$$$$Hencetheareais=2\times \frac{1}{2}\times 2\times 2\sqrt{3}=4\sqrt{3}$$

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