The area of the polygon, whose vertices are the non-real roots of the equation z̅ = iz² is :

Solution

Maths sir · 5 steps

Compare moduli first to get |z| = 1, then z̅ = 1/z turns it into a cube-root equation.

1. Take moduli. Taking modulus on both sides gives |z| = |z|², so |z| = 0 or |z| = 1. The root z = 0 is real, so it is not a vertex. Every non-real root lies on the unit circle.
2. Use the unit circle. On |z| = 1 we have z̅ = 1/z. The equation becomes 1/z = iz², that is z³ = 1/i = -i.
3. Find the roots. Write -i = e^(-iπ/2). The cube roots have arguments -π/6, π/2, 7π/6, giving z = i, √3/2 - i/2, -√3/2 - i/2. All three are non-real, and z = i checks out: z̅ = -i and i · i² = -i.
4. Equilateral triangle. The three points are 120° apart on a circle of radius 1, so they form an equilateral triangle of side √3. Its area is √3/4(√3)².

   Area = √3/4 × 3 = 3√3/4

5. So. The polygon is an equilateral triangle of area 3√3/4, which is option (a).

The idea to keep. When z̅ and powers of z appear together, compare moduli first. Once |z| = 1, replace z̅ with 1/z and you get roots evenly spaced on a circle.

Answer from the NTA final answer key. Solution by PrepWiser.