If z is a complex number such that |z| ≤ 1, then the minimum value of |z+1/2(3+4 i)| is :

Solution

Maths sir · 4 steps

Read the modulus as a distance: how close can a point in the unit disc get to the point -3/2-2i?

1. Turn it into distance. Write z + 1/2(3+4i) = z - (-3/2 - 2i). So the expression is the distance from z to the fixed point P = (-3/2, -2).
2. Locate both pieces. |z| ≤ 1 means z lies inside or on the unit circle centred at the origin. The point P is at distance √(9/4 + 4) = 5/2 from the origin, so it lies outside that disc.

   |z + 1/2(3+4i)| ≥ |1/2(3+4i)| - |z| ≥ 5/2 - 1 = 3/2

3. Check it is reached. The closest point of the disc lies on the line joining the origin to P, at z = -(3+4i)/5. There z + 1/2(3+4i) = 3/10(3+4i), whose modulus is 3/10 × 5 = 3/2.
4. So. The minimum value is 3/2, which is option (c).

The idea to keep. |z - a| is the distance from z to a. For |z| ≤ r and |a| > r, the nearest distance is |a| - r and the farthest is |a| + r.

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