Let $a, b, x$ and $y$ be real numbers such that $a-b=1$ and $y
eq 0$. If the complex number $z=x+i

Question

Let $a, b, x$ and $y$ be real numbers such that $a-b=1$ and $y 
eq 0$. If the complex number $z=x+i y$ satisfies $\operatorname{Im}\left(\frac{a z+b}{z+1}ight)=y$, then which of the following is (are) possible value (s) of $x$ ?\

Professor PC Thomas Classes · Answer

(b, d) : $\operatorname{Im}\left(\frac{a(x+i y)+b}{x+i y+1}ight)=y$\ $\Rightarrow \operatorname{Im}\left(\frac{(a x+b)+i a y}{x+1+i y}ight)=y$\ $\Rightarrow \operatorname{Im}\left(\frac{[(a x+b)+i a y][(x+1)-i y]}{(x+1)^{2}+y^{2}}ight)=y$\ $\Rightarrow \operatorname{Im}\binom{\frac{(a x+b)(x+1)+a y^{2}}{(x+1)^{2}+y^{2}}}{+i \frac{a y(x+1)-y(a x+b)}{(x+1)^{2}+y^{2}}}=y$ Then we have, $\frac{a y-b y}{(x+1)^{2}+y^{2}}=y$\ $\left.\Rightarrow \quad y=y\left\{(x+1)^{2}+y^{2}ight)ight\}$\ As $y 
eq 0$, we have $(x+1)^{2}+y^{2}=1$\ $\Rightarrow \quad x+1= \pm \sqrt{1-y^{2}}$\ $	herefore x=-1+\sqrt{1-y^{2}}$ or $-1-\sqrt{1-y^{2}}$\