Area under the curveMedium

Question

One vertex of the triangle of maximum area that can be inscribed in the curve |z - 2i| = 2 is 2 + 2i, remaining vertices can be -

Options

A.-1+ i(2 + √3)
B.-1- i(2 + √3)
C.-1 + i(2 - √3)
D.-1 - i(2 - √3)

Solution

           
Clearly, the triangle is equilateral. using rotation concept

z1 - 2i = [(2+ 2i) - 2i].e
∴ ahnd z - 2i = [(2 + 2i) - 2i].e
∴ z2 = - 1 + i(2 + √), z2 = - 1+ i(2 - √3)

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