Question

# Let $$\displaystyle z_{1}$$,$$\displaystyle z_{2}$$,$$\displaystyle z_{3}$$are vertices of an equilateral triangle circumscribing the circle  $$\left | z \right |=1$$,if $$\displaystyle z_{1}=1+i\sqrt{3}$$ and$$\displaystyle z_{1}$$,$$\displaystyle z_{2}$$,$$\displaystyle z_{3}$$ are in anticlockwise sense, then $$z_2$$ equals

A
2
B
2
C
i+32
D
1+i32

Solution

## The correct option is A $$-2$$This is a circle with centre at origin and radius $$=1$$$$|z| =1$$ .$$z_{1} = 1+i\sqrt 3 = 2(cos \dfrac{\pi}{3} + isin\dfrac{\pi}{3}) = 2e^{i\pi /3}$$From figure, $$z_{2}$$ can be obtained by rotating $$z_{1}$$ by $$120^{\circ}$$$$\therefore z_{2} = z_{1} e^{i2\pi /3} = 2e^{\pi} = 2(cos \pi +isin\pi) = -2$$Maths

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