Question

The range of real number a for which the equation z + a |z – 1| + 2i = 0 has a solution is –

• A. $\left\lbrack –\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2} \right\rbrack$
• B. $\left\lbrack –\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2} \right\rbrack$
• C. $\left\lbrack 0,\frac{\sqrt{5}}{2} \right\rbrack$
• D. $\left( –\infty,–\frac{\sqrt{5}}{2} \right\rbrack$Č$\left\lbrack \frac{\sqrt{5}}{2},\infty \right)$

Answer 1

Answer: (A)

$\left\lbrack –\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2} \right\rbrack$

Answer 2

Sol. Let z = x + iy.

We have, z + a|z – 1| + 2i = 0

Ž x + i (y + 2) + a$\sqrt{(x - 1)^{2} + y^{2}}$= 0

Equating real and imaginary parts

y + 2 = 0 \ y = –2 and x + a$\sqrt{(x - 1)^{2} + 4}$= 0

\ x² = a² (x² – 2x + 5)

or (1 – a²)x² + 2a²x – 5a² = 0

Since x is real,

\ D = B² – 4aC ³ 0

Ž 4a⁴ + 20a² (1 – a²) ³ 0

Ž –4a⁴ + 5a² £ 0

Ž 4a² $\left( \alpha^{2} - \frac{5}{4} \right)$ £ 0

Ž $\frac{- \sqrt{5}}{2}$ £ a £ $\frac{\sqrt{5}}{2}$.