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Question Number 19687 by Tinkutara last updated on 14/Aug/17

Let z_1 , z_2 , z_3 be three vertices of an equilateral triangle circumscribing the circle ∣z∣ = (1/2). If z_1 = (1/2) + (((√3)i)/2) and z_1 , z_2 , z_3 are in anticlockwise sense then z_2 is

$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{three}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{circumscribing}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mid{z}\mid\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{If}\:{z}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\:\mathrm{and}\:{z}_{\mathrm{1}} , \\ $$$${z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{anticlockwise}\:\mathrm{sense}\:\mathrm{then}\:{z}_{\mathrm{2}} \:\mathrm{is} \\ $$

Answered by ajfour last updated on 14/Aug/17

z_2 =z_1 e^(i2π/3) =e^(iπ/3) .e^(i2π/3) =e^(iπ) =−1 .

$$\mathrm{z}_{\mathrm{2}} =\mathrm{z}_{\mathrm{1}} \mathrm{e}^{\mathrm{i2}\pi/\mathrm{3}} =\mathrm{e}^{\mathrm{i}\pi/\mathrm{3}} .\mathrm{e}^{\mathrm{i2}\pi/\mathrm{3}} =\mathrm{e}^{\mathrm{i}\pi} =−\mathrm{1}\:. \\ $$

Commented by Tinkutara last updated on 14/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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