   Question Answers

# Let ${{z}_{1}}$ and ${{z}_{2}}$ be complex numbers such that ${{z}_{1}}\ne {{z}_{2}}$ and $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|$ . If $\operatorname{Re}\left( {{z}_{1}} \right)>0$ and $\operatorname{Im}\left( {{z}_{2}} \right)<0$ , then $\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ isa. oneb. real and positivec. real and negatived. purely imaginary  Hint: When the modulus of two complex numbers is the same then they are either equal or conjugate of each other. In the given problem they are not equal so they are conjugate. The modulus of two complex numbers may be equal but the numbers may not be equal.

A complex number has two parts, real and imaginary. These represent the x and y coordinates of the point being represented by the complex number. The modulus of a complex number $z$ is given by $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ . Here, we have been given with four conditions for such numbers ${{z}_{1}}\ne {{z}_{2}}$,$\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|$ , $\operatorname{Re}\left( {{z}_{1}} \right)>0$ and $\operatorname{Im}\left( {{z}_{2}} \right)<0$.This means the number ${{z}_{2}}$ is conjugate of number ${{z}_{1}}$ .

Let’s assume the numbers as follow,
${{z}_{1}}=a+ib............(i)$
${{z}_{2}}=a-ib.......(ii)$

Calculating $\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ by substituting the values from equation (i) and equation (ii), we get
\begin{align} & \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{\left( a+ib \right)+\left( a-ib \right)}{\left( a+ib \right)-\left( a-ib \right)} \\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{a+ib+a-ib}{a+ib-a+ib} \\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{2a}{2ib} \\ \end{align}

Above equation can be represented as,
$\Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{1\times a}{ib}...............(iii)$
As, we know ${{i}^{2}}=-1$ then, $1={{\left( -1 \right)}^{2}}={{\left( {{i}^{2}} \right)}^{2}}={{\iota }^{4}}$

Substituting the values in equation (iii), we get
\begin{align} & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{{{i}^{4}}\times a}{ib} \\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{{{i}^{3}}\times a}{b} \\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{i({{i}^{2}})\times a}{b} \\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{-a}{b}i \\ \end{align}

Hence, the number we got is negative and purely imaginary.

Therefore, $\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is negative and purely imaginary number. This means the final answer is option (d).

Note: The chances of mistakes are if the conjugate of number is not taken correctly or there may be mistakes during the interpretation of the conditions given in the question. One should not get confused with the modulus and the number.

Perfect Numbers  Prime Numbers  Cardinal Numbers  Properties of Whole Numbers  CBSE Class 7 Maths Chapter 9 - Rational Numbers Formulas  CBSE Class 8 Maths Chapter 1 - Rational Numbers Formulas  CBSE Class 6 Maths Chapter 2 - Whole Numbers Formulas  CBSE Class 10 Maths Chapter 1 - Real Numbers Formula  CBSE Class 6 Maths Chapter 3 - Playing with Numbers Formulas  CBSE Class 8 Maths Chapter 16 - Playing with Numbers Formulas  Important Questions for CBSE Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations    Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers  Important Questions for CBSE Class 6 Maths Chapter 2 - Whole Numbers  Important Questions for CBSE Class 8 Maths Chapter 1 - Rational Numbers  Important Questions for CBSE Class 8 Maths Chapter 16 - Playing with Numbers  Important Questions for CBSE Class 6 Maths Chapter 1 - Knowing Our Numbers  Important Questions for CBSE Class 7 Maths Chapter 9 - Rational Numbers  Important Questions for CBSE Class 6 Maths Chapter 3 - Playing with Numbers  Important Questions for CBSE Class 11 Maths Chapter 7 - Permutations and Combinations  CBSE Class 12 Maths Question Paper 2020  CBSE Class 10 Maths Question Paper 2020  CBSE Class 10 Maths Question Paper 2017  Maths Question Paper for CBSE Class 10 - 2011  Maths Question Paper for CBSE Class 10 - 2008  Maths Question Paper for CBSE Class 10 - 2012  Maths Question Paper for CBSE Class 10 - 2009  Maths Question Paper for CBSE Class 10 - 2010  Maths Question Paper for CBSE Class 10 - 2007  Maths Question Paper for CBSE Class 12 - 2013  NCERT Solutions for Class 11 Maths Chapter 5  RD Sharma Class 11 Maths Solutions Chapter 13 - Complex Numbers  NCERT Exemplar for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations (Book Solutions)  RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations  RD Sharma Class 11 Solutions Chapter 13 - Complex Numbers (Ex 13.4) Exercise 13.4  NCERT Solutions for Class 11 Maths Chapter 5-Complex Numbers and Quadratic Equations Exercise 5.3  NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations (Ex 5.1) Exercise 5.1  NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations in Hindi  RD Sharma Class 11 Solutions Chapter 13 - Complex Numbers (Ex 13.1) Exercise 13.1  RD Sharma Class 11 Solutions Chapter 13 - Complex Numbers (Ex 13.3) Exercise 13.3  