Complex Numbers 1 Question 10
10. Let $a, b, x$ and $y$ be real numbers such that $a-b=1$ and $y \neq 0$. If the complex number $z=x+i y$ satisfies $\operatorname{Im} \frac{a z+b}{z+1}=y$, then which of the following is(are) possible value(s) of $x$ ?
(2017 Adv.)
(a) $1-\sqrt{1+y^{2}}$
(b) $-1-\sqrt{1-y^{2}}$
(c) $1+\sqrt{1+y^{2}}$
(d) $-1+\sqrt{1-y^{2}}$
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Answer:
Correct Answer: 10. (b, d)
Solution:
- $\frac{a z+b}{z+1}=\frac{a x+b+a i y}{(x+1)+i y}=\frac{(a x+b+a i y)((x+1)-i y)}{(x+1)^{2}+y^{2}}$
$\therefore \quad \operatorname{Im} \frac{a z+b}{z+1}=\frac{-(a x+b) y+a y(x+1)}{(x+1)^{2}+y^{2}}$
$$ \begin{aligned} & \Rightarrow \quad \frac{(a-b) y}{(x+1)^{2}+y^{2}}=y \\ & \because \quad a-b=1 \\ & \therefore \quad(x+1)^{2}+y^{2}=1 \\ & \therefore \quad x=-1 \pm \sqrt{1-y^{2}} \end{aligned} $$
