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 $Im\left(\frac{az+b}{z+1}\right)=y$, 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: (A), (B)

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

$a-b=1$ and $z=x+iy$
$\left(\frac{az+b}{z+1}\right)=\frac{(ax+b)+i\left(ay\right)}{(x+1)+iy}=\frac{\left\{\right(ax+b)+i(ay\left)\right\}\left\{\right(x+1)-iy\}}{(x+1{)}^2+y^2}$
$Im\left(\frac{az+b}{z+1}\right)=\frac{ay(x+1)-y(ax+b)}{(x+1{)}^2+y^2}\Rightarrow y=\frac{ay(x+1)-y(ax+b)}{(x+1{)}^2+y^2}\Rightarrow y=\frac{y(a-b)}{(x+1{)}^2+y^2}\Rightarrow 1=\frac{1}{(x+1{)}^2+y^2}$
$x=-1\pm \sqrt{1-y^2}$