Let a circle C in complex plane pass through the points z₁ = 3 + 4i, z₂ = 4 + 3i and z₃ = 5i. If z( ≠ z₁) is a point on C such that the line through z and z₁ is perpendicular to the line through z₂ and z₃, then arg(z) is equal to :

Solution

Maths sir · 5 steps

All three points have modulus 5, so the circle is just |z| = 5. Now it is plain coordinate geometry.

1. Spot the circle. |3+4i| = |4+3i| = |5i| = 5. All three points are 5 from the origin, so C is the circle x² + y² = 25.
2. Perpendicular line. The line through z₂ = (4, 3) and z₃ = (0, 5) has slope (5-3)/(0-4) = -1/2. A line perpendicular to it has slope 2, so the line through z₁ = (3, 4) is y = 2x - 2.
3. Meet the circle. Put y = 2x - 2 into x² + y² = 25. This gives 5x² - 8x - 21 = 0, and its roots are x = 3 and x = -7/5.

   x = -7/5, y = 2(-7/5) - 2 = -24/5 ⇒ z = -7/5 - 24/5i

4. Read the argument. The root x = 3 is z₁ itself, so z = -7/5 - 24/5i. Both parts are negative, so z is in the third quadrant with tan θ = 24/7.
5. So. For a third-quadrant point the principal argument is tan⁻¹(24/7) - π, which is option (b).

The idea to keep. Before any rotation formula, check the moduli. If the given points share a modulus, the circle is centred at the origin, and the problem turns into simple coordinate geometry.

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