If z=1/2-2 i is such that |z+1|=α z+β(1+i), i=√-1 and α, β ∈ ℝ, then α+β is equal to

Solution

Maths sir · 4 steps

A modulus is real, so the imaginary part on the right must be zero. That links β to α.

1. Find the modulus. z + 1 = 3/2 - 2i, so |z+1| = √(9/4 + 4) = √(25/4) = 5/2. This is a real number.
2. Expand the right side. α and β are real, so split α z + β(1+i) into its real and imaginary parts.

   α z + β(1+i) = (α/2 + β) + i(β - 2α)

3. Compare parts. Imaginary parts: β - 2α = 0, so β = 2α. Real parts: α/2 + 2α = 5/2, so 5α/2 = 5/2. This gives α = 1 and β = 2.
4. So. α + β = 1 + 2 = 3, which is option (c).

The idea to keep. A modulus is always real. When it equals a complex expression, set the imaginary part to zero and match the real part: two equations for two real unknowns.

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