**Miscellaneous Exercise**

Complex Numbers and Quadratic Numbers

**Question and Answers**

**Class 11 – Maths**

Complex Numbers and Quadratic Numbers

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Complex Numbers and Quadratic Numbers |

Chapter No. | Chapter 4 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question 1 Evaluate **

**Answer **

**Question 2 For any two complex numbers z _{1} and z _{2}, prove that**

**Re (z _{1}z_{2})_{ }= Re z_{1 }Re z_{2} – Im z_{1} Im z_{2}**

**Answer**

Let Let z_{1}=x_{1}+iy_{1} and z_{2}=x_{2}+iy_{2
}Product of these two complex numbers, z_{1} z_{2
}z_{1} z_{2 }= ( x_{1}+iy_{1} ) (x_{2}+iy_{2})

z_{1} z_{2 }= x_{1} (x_{2}+iy_{2}) + iy_{1 }(x_{2}+iy_{2})

z_{1} z_{2 }= x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x_{2}+i^{2}y_{1}y_{2
}z_{1} z_{2 }= x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x^{2}−y_{1}y_{2
}z_{1} z_{2 }= (x_{1}x_{2}−y_{1}y_{2})+i(x_{1}y^{2}+y_{1}x^{2})

Re(z_{1}z_{2})=x_{1}x_{2}−y_{1}y_{2
}Re (z_{1}z_{2})_{ }= Re z_{1 }Re z_{2} – Im z_{1} Im z_{2}

**Question 3 Reduce to the standard form**

**Answer**

**Question 4**

**Answer**

**Question 5 If z _{ 1} = 2 – i, z _{2} = 1 + i, find**

**Answer**

**Question 6**

**Answer**

**Question 7 Let z _{1} = 2 – i, z_{2} = –2 + i. Find**

**(i) **_{}

z_{1} = 2 – i, z_{2} = –2 + i

z_{1} z_{2} = (2-i) (-2 +1)

z_{1} z_{2}= −4+2i+2i−i^{2
}z_{1} z_{2 }= −4+4i−(−1)

z_{1} z_{2}= -3 + 4i

**(ii) **_{}

**Question 8 Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i**

**Answer**

z = (*x* – *iy*) (3 + 5*i*)

z = 3x + 5 xi− 3 yi− 5yi^{2
}z = 3x + 5xi − 3yi + 5y

z = (3x+5y) + i (5x−3y)

z̅ *= *− 6 − 24i

(3x+5y) + i (5x−3y) = − 6 − 24i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of *x* in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of *x *and *y* are 3 and –3 respectively.

**Question 9 Find the modulus of**

**Answer**

**Question 10 If (x + iy) ^{3} = u + iv, then show that **

**Answer**

(x + iy)^{3} = u +iv

x^{3 }+ (iy)^{3}+ 3 × x × iy (x+iy) = u+iv

x^{3 }+ i^{3}y^{3 }+ 3x^{2}yi + 3xy^{2}i^{2} = u+iv

x^{3}−iy^{3}+3x^{2}yi−3xy^{2}=u+iv

(x^{3}−3xy^{2})+i(3x^{2}y−y^{3})=u+iv

On equating real and imaginary

u=x^{3}−3xy^{2 },v=3x^{2}y−y^{3}

= x^{2}−3y^{2}+3x^{2}−y^{2
}=4x^{2} -4y^{2
}=4 (x^{2}– y^{2})

**Question 11 If α and β are different complex numbers with |β| = 1, then find**

**Answer**

Let α = a + ib and β = x +iy

|β|= 1

**Question 12) Find the number of non-zero integral solutions of the equation |1 – i| ^{x} = 2^{x}**

**Answer
**|1 – i|

^{x}= 2

^{x}

2^{x/2} = 2^{x}

x/2 = x

x = 2x

2x – x = 0

x = 0

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non- zero integral solutions of the given equation is 0.

**Question 13 If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that
(a ^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2} )(g^{2} + h^{2}) = A^{2} + B^{2}. **

**Answer**

(a + ib)(c + id)(e + if)(g + ih) = A + iB

∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B| **[|z _{1} z_{2}| z_{1}||z_{2}|] **

On squaring both sides, we obtain

(a^{2} + b^{2} )(c^{2} + d^{2} )(e^{2} + f^{2} )(g^{2} + h^{2} ) = A^{2} + B^{2} .

Hence, proved.

**Question 14** **If [(1 +i)/(1 – i)] ^{m} = 1, then find the least positive integral value of m. **

**Answer **

i^{m} =1

i^{m}= i^{4k}

m=4k where k is some integer. Therefore, the least positive is one. Thus, the least positive integral value of m is 4=( 4× 1)

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