Let α = 8 - 14i,A = { z ∈ c:(α z - α̅ z̅)/(z² - ( z̅ )² - 112i)=1 } and B = { z ∈ c:| z + 3i | = 4 }. Then Σ_(z ∈ A ∩ B) ( Re z - Im z) is equal to ____________.

Solution

Maths sir · 4 steps

Put z = x + iy and the messy fraction collapses into two straight lines that each touch the circle once.

1. Simplify the fraction. With z = x + iy, the top is α z - α̅ ̅z̅ = 2i Im(α z) = 2i(8y - 14x). The bottom is z² - z̄² - 112i = i(4xy - 112), so the i cancels.

   2(8y - 14x)/(4xy - 112) = 1 ⇒ xy + 7x - 4y - 28 = 0 ⇒ (x - 4)(y + 7) = 0

2. Set A. So A is the pair of lines x = 4 and y = -7, leaving out points where xy = 28, because there the denominator is zero.
3. Meet the circle. B is x² + (y + 3)² = 16, centre (0, -3), radius 4. Putting x = 4 forces (y + 3)² = 0, so the point is (4, -3). Putting y = -7 forces x² = 0, so the point is (0, -7). Neither point has xy = 28, so both are allowed.
4. So. The sum is (4 - (-3)) + (0 - (-7)) = 7 + 7 = 14.

The common slip: 7. Counting only one intersection point gives 7. Both lines touch the circle, so there are two points, each contributing 7.

The idea to keep. Expressions like α z - ᾱz̄ and z² - z̄² are always pure imaginary. Write z = x + iy, cancel the i, and look for a factorisation into lines.

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