Question

\( z \) be a complex number satisfying \( |z-5i|\le 1 \) such that \( \arg z \) is minimum, then \( z = \)

• A. \( 1+i2\sqrt{6} \)
• B. \( \dfrac{1+i2\sqrt{6}}{5} \)
• C. \( \dfrac{2\sqrt{6}}{5}(1+i2\sqrt{6}) \)
• D. \( \dfrac{2\sqrt{6}}{5}(1-i2\sqrt{6}) \)

Hint:

For problems involving extremum of \( \arg z \) under a circular constraint, draw tangents from the origin to the circle and use the point of contact.

Answer 1

The Correct Option is C

The condition \( |z-5i|\le 1 \) represents a circle with centre \( (0,5) \) and radius \( 1 \) in the Argand plane. Step 1: To make \( \arg z \) minimum, draw a tangent from the origin to the circle. Let the tangent make an angle \( \theta \) with the positive real axis, so its equation is \[ y = mx \quad \text{where } m=\tan\theta. \]
Step 2: Distance of the centre \( (0,5) \) from the line \( y=mx \) is equal to the radius: \[ \frac{5}{\sqrt{m^2+1}} = 1 \] \[ \Rightarrow m^2+1 = 25 \Rightarrow m^2 = 24 \Rightarrow m = 2\sqrt{6}. \] Thus, the tangent is \( y = 2\sqrt{6}x \).
Step 3: The point of contact lies on both the line and the circle: \[ x^2 + (2\sqrt{6}x - 5)^2 = 1. \] Solving, \[ x = \frac{2\sqrt{6}}{5}, \quad y = \frac{24}{5}. \]
Step 4: Hence, \[ z = \frac{2\sqrt{6}}{5} + i\frac{24}{5} = \frac{2\sqrt{6}}{5}(1+i2\sqrt{6}). \]