Question

Find the range of real number $\alpha$ for which the equation $z+\alpha |z-1|+2i=0;z=x+iy$ has a solution. Find the solution.

• A. $x=5/2,y=-2$
• B. $x=-2,y=5/2$
• C. $x=-5/2,y=2$
• D. $x=2,y=-5/2$

Answer 1

The correct option is A $x=5/2,y=-2$

$x+iy+\alpha \left(\sqrt{(x-1{)}^2+y^2}\right)+2i=0$
$x+i(y+2)=\alpha (\sqrt{(x-1{)}^2+y^2}$
$x^2-(y+2{)}^2+i2x(y+2)={\alpha }^2((x-1{)}^2+y^2)$
Real part
$x^2-\alpha (x-1{)}^2-\alpha (y^2)=0$ and
Imaginary part
$2x(y+2)=0$
$x(y+2)=0$
Either $x=0$or $y=-2$ ...if the above equations have a solution.
Considering y=-2,
Hence
$\alpha ϵ[-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}]$
Hence
$x=\frac{5}{2}$ if $|\alpha |\ne 1$.