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 $\frac{(az+b)}{(z+1)}=y$, then which of the following is (are) possible value(s) of x?

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

Answer 1

The correct option is B ($-1-\sqrt{1-y^2}$) & ($-1+\sqrt{1-y^2}$)

$\frac{az+b}{z+1}=\frac{ax+b+aiy}{(x+1)+iy}=\frac{(ax+b+aiy)(x+1)-iy}{(x+1{)}^2+y^2}\therefore Im\frac{az+b}{z+1}=\frac{-(ax+b)y+ay(x+1)}{(x+1{)}^2+y^2}\Rightarrow \frac{(a-b)y}{(x+1{)}^2+y^2}=y\because a-b=1\therefore (x+1{)}^2+y^2=1\therefore x=-1\pm \sqrt{1-y^2}$