Question

Let $a,b,x$ and $y$ be real numbers such that $a–b=1$ and $y\ne 0$. If the complex number $z=x+iy$ satisfies $lm\left(\frac{az+b}{z+1}\right)=y$ , then which of the following is(are) possible value(s) of $x$ ?

• A. $1-\sqrt{1+y^2}$
• B. $-1-\sqrt{1-y^2}$
• C. $-1+\sqrt{1-y^2}$
• D. $1+\sqrt{1+y^2}$

Answer 1

Answer: (B), (C)

$-1+\sqrt{1-y^2}$

$a-b\ne 1,y\ne 0,z=x+iy$
$lm\left(\frac{az+b}{z+1}\right)=y$
$lm(\left(\frac{(ax+b)+ayi}{(x+1)+iy}\right)\times \left(\frac{x+1-iy}{x+1-iy}\right))=y$
$lm\left(\frac{(ax+b)(x+1)+a{y}^2+ay(z+1)i-iy(ax+b)}{(x+1{)}^2+y^2}\right)=y$
$\Rightarrow ay(x+1)-y(ax+b)=y(x+1{)}^2+y^3$
$\Rightarrow ax+a-ax-b=x^2+2x+1+y^2$
$\Rightarrow a-b=x^2+y^2+2x+1$
$\Rightarrow 1=x^2+y^2+2x+1$
$\Rightarrow (x+1{)}^2=1-y^2$
$\Rightarrow x+1=\pm \sqrt{1-y^2}$
$\Rightarrow x=-1\pm \sqrt{1-y^2}$