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

What is the area bounded by the curves arg(z) = (π/3) ; arg(z)= ((2π)/3) and arg(z−2−2i(√3))=π on the complex plane?

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

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

(i) arg (z) = (π/3) → (y/x) = tan ((π/3)) y = x(√3) (ii) arg (z) = ((2π)/3)→(y/x) = tan (((2π)/3)) y = −x(√3) (iii) arg (z−2−2i(√3) )= π ((y−2(√3))/(x−2)) = tan π →y = 2(√3) Hence the area is = 2×(1/2)×2×2(√3) = 4(√3)

$$\:\left({i}\right)\:{arg}\:\left({z}\right)\:=\:\frac{\pi}{\mathrm{3}}\:\rightarrow\:\frac{{y}}{{x}}\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:{y}\:=\:{x}\sqrt{\mathrm{3}}\: \\ $$$$\:\:\left({ii}\right)\:{arg}\:\left({z}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\rightarrow\frac{{y}}{{x}}\:=\:\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:{y}\:=\:−{x}\sqrt{\mathrm{3}} \\ $$$$\:\left({iii}\right)\:{arg}\:\left({z}−\mathrm{2}−\mathrm{2}{i}\sqrt{\mathrm{3}}\:\right)=\:\pi\: \\ $$$$\:\:\:\frac{{y}−\mathrm{2}\sqrt{\mathrm{3}}}{{x}−\mathrm{2}}\:=\:\mathrm{tan}\:\pi\:\rightarrow{y}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${Hence}\:{the}\:{area}\:{is}\:=\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}\:=\:\mathrm{4}\sqrt{\mathrm{3}} \\ $$

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