Answer
Verified
426.6k+ views
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 $z=x+iy$
$\begin{align}
& \Rightarrow \dfrac{a\left( x+iy \right)+b}{x+iy+1} \\
& \Rightarrow \dfrac{\left( ax+b \right)+iya}{\left( x+1 \right)+iy} \\
\end{align}$
Now, we need to rationalize the above expression to make denominator real; Hence multiplying with conjugate of denominator in whole fraction a following:
\[\begin{align}
& =\dfrac{\left( ax+b \right)iya}{\left( x+y \right)+iy}\times \left( \dfrac{\left( x+1 \right)-iy}{\left( x+1 \right)-iy} \right) \\
& =\dfrac{\left( \left( ax+b \right)\left( x+1 \right)-{{i}^{2}}ya \right)+iya\left( x+1 \right)-iy\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}-{{i}^{2}}{{y}^{2}}-iy\left( x+1 \right)+iy\left( x+1 \right)} \\
& =\dfrac{\left( ax+1 \right)\left( x+1 \right)+ya+i\left( ya\left( x+1 \right)-y\left( ax+b \right) \right)}{{{\left( x+1 \right)}^{2}}+y} \\
& \left( {{i}^{2}}=-1 \right) \\
\end{align}\]
Now,
\[\begin{align}
& \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ay\left( x+1 \right)-y\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}} \\
& \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ayx+ya-ayx-yb}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}...........\left( iii \right) \\
\end{align}\]
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;
$\begin{align}
& {{\left( x+1 \right)}^{2}}+{{y}^{2}}=1 \\
& {{\left( x+1 \right)}^{2}}=1-{{y}^{2}} \\
& \left( x+1 \right)=\pm \sqrt{1-{{y}^{2}}}\text{ if }{{\text{N}}^{2}}=X\Rightarrow N=\pm \sqrt{X} \\
& x=-1\pm \sqrt{1-{{y}^{2}}} \\
\end{align}$
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,
\[\begin{align}
& z=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
& \overline{z}=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
\end{align}\]
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 $z=x+iy$
$\begin{align}
& \Rightarrow \dfrac{a\left( x+iy \right)+b}{x+iy+1} \\
& \Rightarrow \dfrac{\left( ax+b \right)+iya}{\left( x+1 \right)+iy} \\
\end{align}$
Now, we need to rationalize the above expression to make denominator real; Hence multiplying with conjugate of denominator in whole fraction a following:
\[\begin{align}
& =\dfrac{\left( ax+b \right)iya}{\left( x+y \right)+iy}\times \left( \dfrac{\left( x+1 \right)-iy}{\left( x+1 \right)-iy} \right) \\
& =\dfrac{\left( \left( ax+b \right)\left( x+1 \right)-{{i}^{2}}ya \right)+iya\left( x+1 \right)-iy\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}-{{i}^{2}}{{y}^{2}}-iy\left( x+1 \right)+iy\left( x+1 \right)} \\
& =\dfrac{\left( ax+1 \right)\left( x+1 \right)+ya+i\left( ya\left( x+1 \right)-y\left( ax+b \right) \right)}{{{\left( x+1 \right)}^{2}}+y} \\
& \left( {{i}^{2}}=-1 \right) \\
\end{align}\]
Now,
\[\begin{align}
& \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ay\left( x+1 \right)-y\left( ax+b \right)}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}} \\
& \operatorname{Im}\left( \dfrac{az+b}{z+1} \right)=\dfrac{ayx+ya-ayx-yb}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}...........\left( iii \right) \\
\end{align}\]
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;
$\begin{align}
& {{\left( x+1 \right)}^{2}}+{{y}^{2}}=1 \\
& {{\left( x+1 \right)}^{2}}=1-{{y}^{2}} \\
& \left( x+1 \right)=\pm \sqrt{1-{{y}^{2}}}\text{ if }{{\text{N}}^{2}}=X\Rightarrow N=\pm \sqrt{X} \\
& x=-1\pm \sqrt{1-{{y}^{2}}} \\
\end{align}$
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,
\[\begin{align}
& z=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
& \overline{z}=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n}} \right)+i\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+......{{b}_{n}} \right) \\
\end{align}\]
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE