Question

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

• A. $[-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}]$
• B. $[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}]$
• C. $[0,\frac{\sqrt{5}}{2}]$
• D. $(-\infty ,-\frac{\sqrt{5}}{2}]\cup [\frac{\sqrt{5}}{2},\infty )$

Answer 1

The correct option is A $[-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}]$

We have, $z+\alpha |z-1|+2i=0\Rightarrow x+i(y+2)+\alpha \sqrt{(x-1{)}^2}+y^2=0$
Equating real and imaginary parts
$y+2=0\Rightarrow y=-2$

and $x+\alpha \sqrt{(x-1{)}^2+4}=0$
$\therefore x^2={\alpha }^2(x^2-2x+5)$$\Rightarrow (1-{\alpha }^2)x^2+2{\alpha }^{2}x-5{\alpha }^2=0$
Since $x$ is real,

$\therefore D=B^2-4AC\ge 0\Rightarrow 4{\alpha }^4+20{\alpha }^2(1-{\alpha }^2)\ge 0$

$\Rightarrow -4{\alpha }^4+5{\alpha }^2\ge 0\Rightarrow 4{\alpha }^2({\alpha }^2-\frac{5}{4})\le 0$
Now $\alpha$ being real implies ${\alpha }^2$ is +ve and hence we conclude that ${\alpha }^2-\frac{5}{4}$ i -ve
$\Rightarrow -\frac{\sqrt{5}}{2}\le \alpha \le \frac{\sqrt{5}}{2}$