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 when 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), (D)

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

$Im\left(\frac{(ax+b+iay)(x+1-iy)}{(x+1+iy)(x+1-iy)}\right)=y$

$Im\left(\frac{(ax+b+iay)(x+1-iy)}{(x+1{)}^2+y^2}\right)=y$

$y\left[\right(x+1{)}^2+y^2]=-y(ax+b)+ay(x+1)$ $=-by+ay$ $=(a-b)y=y$

$\therefore y\left[\right(x+1{)}^2+y^2]=y$

$\therefore (x+1{)}^2=1-y^2$

$\therefore (x+1)=\pm \sqrt{1-y^2}$

$x=-1\pm \sqrt{1-y^2}$

Hence, options B and D are correct.