Let z be a complex number such that the real part of (z-2 i)/(z+2 i) is zero. Then, the maximum value of |z-(6+8 i)| is equal to

Solution

Maths sir · 4 steps

Real part zero means z sits on a circle; then the farthest point is distance to centre plus the radius.

1. Purely imaginary condition. A number w has real part zero exactly when w + w̄ = 0. Take w = (z-2i)/(z+2i), so w̄ = (z̄+2i)/(z̄-2i).

   (z-2i)(z̄-2i) + (z̄+2i)(z+2i) = 0 ⇒ 2|z|² - 8 = 0

2. Read the locus. So |z| = 2: a circle with centre 0 and radius 2, leaving out z = -2i where the fraction is undefined. This also follows from geometry: the segment joining 2i and -2i subtends a right angle at z, so z lies on the circle that has this segment as its diameter.
3. Farthest point. |z-(6+8i)| is the distance from z to the point (6, 8). That point is √(36+64) = 10 from the centre, so the largest distance from any point of the circle is 10 + 2. It occurs at z = -1.2 - 1.6i, which is on the circle and is not the excluded point.
4. So. The maximum value of |z-(6+8i)| is 12, option (b).

The idea to keep. For a point moving on a circle, the distance to a fixed outside point ranges from (distance to centre minus radius) to (distance to centre plus radius).

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