Let the minimum value v₀ of v=|z|²+|z-3|²+|z-6 i|², z ∈ ℂ is attained at z=z₀. Then |2 z₀²-z̄₀³+3|²+v₀² is equal to :

Solution

Maths sir · 5 steps

Put z = x + iy, complete the square, and the minimum point and minimum value fall out together.

1. Use coordinates. With z = x + iy, expand each term: v = (x² + y²) + ((x-3)² + y²) + (x² + (y-6)²) = 3x² - 6x + 3y² - 12y + 45.

   v = 3(x-1)² + 3(y-2)² + 30

2. Read off the minimum. Neither square can be negative, so the least value is v₀ = 30, reached at z₀ = 1 + 2i. This point is the centroid of 0, 3 and 6i.
3. Build the expression. z₀² = (1+2i)² = -3 + 4i, so 2z₀² = -6 + 8i. Also z̄₀ = 1 - 2i, z̄₀² = -3 - 4i, and z̄₀³ = (-3-4i)(1-2i) = -11 + 2i.
4. Combine. 2z₀² - z̄₀³ + 3 = (-6 + 8i) - (-11 + 2i) + 3 = 8 + 6i. Its modulus squared is 64 + 36 = 100.
5. So. 100 + v₀² = 100 + 900 = 1000, which is option (a).

The idea to keep. A sum |z - a|² + |z - b|² + |z - c|² is least at the centroid of a, b, c. Complete the square in x and y to get both the point and the minimum value.

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