Question

Let z lies on a circle centred at origin. If area of thetriangle whose vertices are z, az and Z + az, where ais the imaginary cube root of unity, is 4v3 sq. units,then radius of the circle is(2) 2 units(1) 1 unit(3), 3 units(4) 4 unitsno ::​

Answer 1

The radius of the circle is length 4 units.

Step-by-step explanation:

Without loss of generality assume that z = r + 0i .

Since we have that  w = (e^(2π i))^(1/3)    ( using e^(2π i) = 1 + 0i = 1 ) then

w = e^(2π/3i)

In cartesian coordinates, w = cos((2π)/3) + sin((2π)/3)i = -1/2 + √ (3)/2i

so that  z + w.z = r/2 + √ (3)/2 ri

The area of the rhombus is equal to (r/2) x (√ 3)/2r) + 2 x (1/2) x (r/2) x (√3)/2 r).

The two identical triangles at each end.

Which in total is 2 x (r/2) x (√3)/2 r) = √ (3)/2r^2

• The triangle of interest then has area √3/4 r^2 .
• Given that this area is equal to 4√(3) squared units then this implies that r = 4.

Thus the radius of the circle is length 4 units.

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