Let z=1+i and z₁=(1+i z̄)/(z̄(1-z)+1/z). Then 12/π arg (z₁) is equal to __________.

Solution

Maths sir · 4 steps

Put z = 1+i in and simplify to the form a+ib first. Only then look at the quadrant.

1. Numerator. Here z̄ = 1-i, so 1 + iz̄ = 1 + i(1-i) = 1 + i + 1 = 2 + i.
2. Denominator. 1 - z = -i, so z̄(1-z) = (1-i)(-i) = -i + i² = -1 - i. Also 1/z = (1-i)/2. Adding them gives -1 - i + 1/2 - i/2 = -(1+3i)/2.

   z₁ = (2+i)/(-(1+3i)/2) = -2(2+i)(1-3i)/(1+3i)(1-3i) = -2(5-5i)/10 = -1 + i

3. Find the argument. z₁ = -1 + i has a negative real part and a positive imaginary part, so it lies in the second quadrant. Its argument is π - π/4 = 3π/4.
4. So. 12/π × 3π/4 = 9.

The common slip: -3. Taking arg(-1+i) = tan⁻¹(-1) = -π/4 ignores the quadrant and gives -3 instead of 9.

The idea to keep. To find an argument, always mark the quadrant from the signs of the real and imaginary parts first. tan⁻¹(y/x) alone cannot tell the second quadrant apart from the fourth.

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