# 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

364.5k+ 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}\]

Last updated date: 30th Sep 2023

â€¢

Total views: 364.5k

â€¢

Views today: 9.64k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE