Question

Let $𝑎, 𝑏, 𝑥$ and $𝑦$ be real numbers such that $𝑎 − 𝑏 = 1 $ and $𝑦 ≠ 0$. 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)

$\operatorname{Im}\left(\frac{a z+b}{z+1}\right)=y$
$\Rightarrow \operatorname{Im}\left(\frac{a(x+i y)+b}{x+i y+1}\right)=y$
$\Rightarrow \operatorname{Im}\left(\frac{a i y+(a x+b)}{(x+1)+i y} \times \frac{((x+1)-i y)}{((x+1)-i y)}\right)=y$
$ \Rightarrow \frac{a y(x+1)-y(a x+b)}{(x+1)^{2}+y^{2}}=y $
$ \Rightarrow (a-b) y=y\left((x+1)^{2}+y^{2}\right) $
$(x+1)^{2}+y^{2}=1 $
$ x=-1 \pm \sqrt{1-y^{2}} $
$(\because a-b=1) $
${x=-\sqrt{1-y^{2}}} $