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JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Complex Numbers

If

z = \frac{1}{2} - 2 i

, is such that

∣ z + 1∣ = α z + β (1 + i)

,

i = - 1

and

α, β \in R

, then

α + β

is equal to

(A)-4
(B)3✓ Correct
(C)2
(D)-1

Answer: (B)

Solution

z = \frac{1}{2} - 2 i

, so

z + 1 = \frac{3}{2} - 2 i

, giving

∣ z + 1∣ = (\frac{3}{2})^{2} + 2^{2} = \frac{25}{4} = \frac{5}{2}

.

The equation

∣ z + 1∣ = α z + β (1 + i)

becomes

\frac{5}{2} = α (\frac{1}{2} - 2 i) + β (1 + i) = (\frac{α}{2} + β) + i (β - 2 α)

Matching the (real) left side means the imaginary part vanishes:

β = 2 α

.

Real part:

\frac{α}{2} + β = \frac{5}{2} ⟹ \frac{α}{2} + 2 α = \frac{5}{2} ⟹ \frac{5 α}{2} = \frac{5}{2} ⟹ α = 1, β = 2

α + β = 3

Key confirmed: B.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option

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