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Mathematics Question on argand plane

Suppose z1,z2,z3z_1, z_2, z_3 are the vertices of an equilateral triangle inscribed in the circle ∣z∣=2| z| = 2. If z1z_1 = 3i\sqrt {3i} and z1,z2,z3z_1,z_2,z_3 are in . the clockwise sense, then

A

z2=1−3i,z3=−2z_2 = 1 - \sqrt {3i},z_3 =-2

B

z2=,z3=1−3iz_2 = , z_3 = 1 - \sqrt {3i}

C

z2=−1+3i,z3=−2z_2 = -1 + \sqrt {3i},z_3 =-2

D

none of these

Answer

z2=1−3i,z3=−2z_2 = 1 - \sqrt {3i},z_3 =-2

Explanation

Solution

amp.(z1)=amp.(1+3i)amp. \left(z_{1}\right)=amp. \left(1+\sqrt{3}i\right) =tan−1 3=π3=tan^{-1}\,\sqrt{3}=\frac{\pi}{3} amp.(z2)=π3−2π3=−π3amp. \left(z_{2}\right)=\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3} and ∣z2∣=2\left|z_{2}\right|=2 ∴z2=2[cos(−π3)+i sin(−π3)]\therefore z_{2}=2\left[cos\left(-\frac{\pi}{3}\right)+i\,sin \left(-\frac{\pi}{3}\right)\right] =2[cosπ3−i sinπ3]=2\left[cos \frac{\pi}{3}-i\,sin \frac{\pi}{3}\right] =2(12−i32)=1−i3=2\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=1-i\sqrt{3} amp.(z3)=π3+2π3=πamp. \left(z_{3}\right)=\frac{\pi}{3}+\frac{2\pi}{3}=\pi and ∣z3∣=2\left|z_{3}\right|=2 ∴z3=2(cos π+i sin π)=2\therefore z_{3}=2\left(cos\,\pi+i\,sin\,\pi\right)=2