Let $a,b,x\text{ and }y$ be real numbers such that $a-b=1\text{ and }y\ne 0.$ If the complex number $z=x+iy$ satisfies $\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=y,$ 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 Verified Verified
Hint: Rationalize $\left( \dfrac{az+b}{z+1} \right)$ after putting $z=x+iy$ then use the given condition.

We have given that $a,b,x\text{ and }y$ are real numbers with
$a-b=1\text{ }..............\left( i \right)$
$\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=y.............\left( ii \right)$
Where $z=x+iy$
As we know that any complex number $w=x+iy$ has two parts.$x$ is real part and $y$ is imaginary part and every complex number can be written in form of $x+iy.$
So, let us calculate
$\left( \dfrac{az+b}{z+1} \right)$ from equation (ii) by putting $z=x+iy$
  & \Rightarrow \dfrac{a\left( x+iy \right)+b}{x+iy+1} \\
 & \Rightarrow \dfrac{\left( ax+b \right)+iya}{\left( x+1 \right)+iy} \\
Now, we need to rationalize the above expression to make denominator real; Hence multiplying with conjugate of denominator in whole fraction a following:
  & =\dfrac{\left( ax+b \right)iya}{\left( x+y \right)+iy}\times \left( \dfrac{\left( x+1 \right)-iy}{\left( x+1 \right)-iy} \right) \\
 & =\dfrac{\left( \left( ax+b \right)\left( x+1 \right)-{{i}^{2}}ya \right)+iya\left( x+1 \right)-iy\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}-{{i}^{2}}{{y}^{2}}-iy\left( x+1 \right)+iy\left( x+1 \right)} \\
 & =\dfrac{\left( ax+1 \right)\left( x+1 \right)+ya+i\left( ya\left( x+1 \right)-y\left( ax+b \right) \right)}{{{\left( x+1 \right)}^{2}}+y} \\
 & \left( {{i}^{2}}=-1 \right) \\
  & \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ay\left( x+1 \right)-y\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}} \\
 & \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ayx+ya-ayx-yb}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}...........\left( iii \right) \\
From equation (ii) and (iii) we have
\[\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{y\left( a-b \right)}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}=y\]
$\dfrac{a-b}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}=1$
As we have $a-b=1$ from equation (i)
  & {{\left( x+1 \right)}^{2}}+{{y}^{2}}=1 \\
 & {{\left( x+1 \right)}^{2}}=1-{{y}^{2}} \\
 & \left( x+1 \right)=\pm \sqrt{1-{{y}^{2}}}\text{ if }{{\text{N}}^{2}}=X\Rightarrow N=\pm \sqrt{X} \\
 & x=-1\pm \sqrt{1-{{y}^{2}}} \\
Hence, (b) and (d) are correct options.

Note: (i) Need not to calculate the real part $\left( \dfrac{az+b}{z+1} \right)$ for the requirement of solution.
(ii) Students can make mistakes with rationalization steps. One can multiply by $\left( x-1+iy \right)$. As students see three numbers in addition, usually questions have $a+ib$ form and need to multiply by $a-ib$ for rationalization. So, we need to place the $'-'$ sign between the real part and imaginary part.
  & z=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
 & \overline{z}=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
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