Question

If the imaginary part of $\frac{2z+1}{iz+1}$ is $-2$, then show that the locus of the point representing $z$ in the argand plane is a straight line.

Answer 1

Solution

Let $z=x+iy$

then $M=\frac{2z+1}{iz+1}=\frac{2x+2+iy}{(1-y)+ix}$

$M=\frac{(2x+1+2iy)\left(\right(1-y)-ix)}{\left(\right(1-y)+ix)\left(\right(1-y)-ix)}$

$M=\frac{(2x+)(1-9)2iy(1-y)-ix(2x+1)+xy}{(1-y{)}^2{x}^2}$

IM $\left(M\right)=\frac{2y(1-y)-x(2x+1)}{(1-y{)}^2+x^2}=-2$

$2y-2y^2-2x^2-x=-2[1+y^2+x^2-2y]$

$2y-x=-2+4y$

$y=\frac{-x}{2}+1$

locus is straight line