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Question: Let \(a,b,x\text{ and }y\) be real numbers such that \(a-b=1\text{ and }y\ne 0.\) If the complex num...

Let a,b,x and ya,b,x\text{ and }y be real numbers such that a−b=1 and y≠0.a-b=1\text{ and }y\ne 0. If the complex number z=x+iyz=x+iy satisfies Im⁡(az+bz+1)=y,\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=y, then when of the following is (are) possible value(s) of xx ?
(a) 1−1+y21-\sqrt{1+{{y}^{2}}}
(b) −1+1−y2-1+\sqrt{1-{{y}^{2}}}
(c) 1+1+y21+\sqrt{1+{{y}^{2}}}
(d) −1−1−y2-1-\sqrt{1-{{y}^{2}}}

Explanation

Solution

Hint: Rationalize (az+bz+1)\left( \dfrac{az+b}{z+1} \right) after putting z=x+iyz=x+iy then use the given condition.

We have given that a,b,x and ya,b,x\text{ and }y are real numbers with
a−b=1 ..............(i)a-b=1\text{ }..............\left( i \right)
Im⁡(az+bz+1)=y.............(ii)\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=y.............\left( ii \right)
Where z=x+iyz=x+iy
As we know that any complex number w=x+iyw=x+iy has two parts.xx is real part and yy is imaginary part and every complex number can be written in form of x+iy.x+iy.
So, let us calculate
(az+bz+1)\left( \dfrac{az+b}{z+1} \right) from equation (ii) by putting z=x+iyz=x+iy
⇒a(x+iy)+bx+iy+1 ⇒(ax+b)+iya(x+1)+iy \begin{aligned} & \Rightarrow \dfrac{a\left( x+iy \right)+b}{x+iy+1} \\\ & \Rightarrow \dfrac{\left( ax+b \right)+iya}{\left( x+1 \right)+iy} \\\ \end{aligned}
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) \\\ \end{aligned}$$ Now, $$\begin{aligned} & \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) \\\ \end{aligned}$$ From equation (ii) and (iii) we have $$\operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{y\left( a-b \right)}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}=y$$ Hence, $\dfrac{a-b}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}=1$ As we have $a-b=1$ from equation (i) Hence; $\begin{aligned} & {{\left( x+1 \right)}^{2}}+{{y}^{2}}=1 \\\ & {{\left( x+1 \right)}^{2}}=1-{{y}^{2}} \\\ & \left( x+1 \right)=\pm \sqrt{1-{{y}^{2}}}\text{ if }{{\text{N}}^{2}}=X\Rightarrow N=\pm \sqrt{X} \\\ & x=-1\pm \sqrt{1-{{y}^{2}}} \\\ \end{aligned}$ Hence, (b) and (d) are correct options. Note: (i) Need not to calculate the real part $\left( \dfrac{az+b}{z+1} \right)$ for the requirement of solution. (ii) Students can make mistakes with rationalization steps. One can multiply by $\left( x-1+iy \right)$. As students see three numbers in addition, usually questions have $a+ib$ form and need to multiply by $a-ib$ for rationalization. So, we need to place the $'-'$ sign between the real part and imaginary part. Let, $$\begin{aligned} & 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) \\\ \end{aligned}$$