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

$(a)\;x+2y-2=0\qquad(b)\;2x+y+1=0\qquad(c)\;2x-2y+1=0\qquad(d)\;x=0$

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Answer : $\;x+2y-2=0$
Explanation :
Let $\; z=x+iy\;$
Then , $\;\large\frac{2z+1}{iz+1}\;= \large\frac{2(x+iy)+1}{i(x+iy) +1}$
$= \large\frac{(2x+1)+i2y }{(1-y)+ix} $
$= \large\frac{(2x+1)+i2y }{(1-y)+ix} \times \large\frac{{(1-y)-ix}}{{(1-y)-ix}}$
$= \large\frac{(2x+y-1) + i(2y-2y^{2}-2x^{2}-x)}{1+y^{2} -2y+x^{2}} $
$\;Im(\large\frac{2z+1}{iz+1})\; = \large\frac{ (2y-2y^{2}-2x^{2}-x)}{1+y^{2} -2y+x^{2}}$
But , given $\;Im(\large\frac{2z+1}{iz+1})\;= -2$
So ,$\;\large\frac{ (2y-2y^{2}-2x^{2}-x)}{1+y^{2} -2y+x^{2}}=-2$
$2y-2y^{2}-2x^{2}-x = -2 (1+y^{2} -2y +x^{2})$
$x+2y-2=0\;$ , which is the equation of a line
answered Apr 17, 2014 by yamini.v
 

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