Let A={z ∈ C:|z-2-i|=3}, B={z ∈ C: Re(z-i z)=2} and S=A ∩ B. Then Σ_(z ∈ S)|z|² is equal to _________.

Solution

Maths sir · 4 steps

Write z = x + iy, cut the circle with the line, and let Vieta add the two |z|² values for you.

1. Read both sets. Put z = x + iy. Set A is the circle (x-2)² + (y-1)² = 9. For B, z - iz = (x + iy) - (ix - y) = (x + y) + i(y - x), so B is the line x + y = 2.
2. Cut circle by line. Put y = 2 - x into the circle: (x-2)² + (1-x)² = 9. This gives 2x² - 6x - 4 = 0, that is x² - 3x - 2 = 0. Its two roots x₁, x₂ give the two points of S, with x₁ + x₂ = 3 and x₁ x₂ = -2.
3. Express |z|² in x. On the line, |z|² = x² + (2 - x)² = 2x² - 4x + 4. Also x₁² + x₂² = (x₁ + x₂)² - 2x₁ x₂ = 9 + 4 = 13.

   Σ_(z ∈ S) |z|² = 2(x₁² + x₂²) - 4(x₁ + x₂) + 8 = 26 - 12 + 8

4. So. The sum of |z|² over both intersection points is 22.

The common slip: 30. Taking Re(z - iz) as x - y instead of x + y puts the wrong line in, and the sum comes out as 30, not 22.

The idea to keep. When you need a symmetric sum over intersection points, do not solve for the roots. Write the quantity in one variable and use the sum and product of roots.

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