Let z1- z2- z3 be three vertices of an equilateral triangle circumscribing the circle -z- - 1-2- If z1 - 1-2 - -3 i-2 and z1- z2- z3 are in anticlockwise sense then z2 is

Let $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $| z |$ = 2. If $z_{1}, z_{2}, z_{3}$ are in the clockwise sense and $z_{1}$ = $1 + i \sqrt{3}$ , then:

Q. The complex numbers

z_{1}, z_{2}

and

z_{3}

satisfying

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1 - \sqrt{3} i}{2}

are the vertices of a triangle which is:

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Let z1,z2,z3 be three vertices of an equilateral triangle circumscribing the circle |z|=12. If z1=12+√3i2 and z1,z2,z3 are in anticlockwise sense then z2 is

The correct option is B -1z2=z1e2iπ/3=(12+√32i)(cos2π3+isin2π3)=(12+√32i)(−12+√32i)=−34−14=−1

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

The correct option is B -1
$z_{2} = z_{1} e^{2 i π / 3} = (\frac{1}{2} + \frac{\sqrt{3}}{2} i) (cos \frac{2 π}{3} + i sin \frac{2 π}{3})$
$= (\frac{1}{2} + \frac{\sqrt{3}}{2} i) (\frac{- 1}{2} + \frac{\sqrt{3}}{2} i)$
$= - \frac{3}{4} - \frac{1}{4} = - 1$