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# Miscellaneous Exercise on Chapter - 5 Class 11 maths 5. Complex Numbers and Quadratic Equations - ncert solutions - Toppers Study

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## Miscellaneous Exercise on Chapter - 5 Class 11 maths 5. Complex Numbers and Quadratic Equations - ncert solutions - Toppers Study

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### Miscellaneous Exercise on Chapter - 5 class 11 Mathematics Chapter 5. Complex Numbers and Quadratic Equations

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- Miscellaneous Exercise On Chapter - 5 Class 11 Maths 5. Complex Numbers And Quadratic Equations - Ncert Solutions - Toppers Study
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- Chapter 5. Complex Numbers And Quadratic Equations Miscellaneous Exercise On Chapter - 5 Class 11

## 5. Complex Numbers and Quadratic Equations

### | Miscellaneous Exercise on Chapter - 5 |

## Miscellaneous Exercise on Chapter - 5 Class 11 maths 5. Complex Numbers and Quadratic Equations - ncert solutions - Toppers Study

**Miscellaneous Exercise on Chapter 5**

**Q2.** **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}**

**Solution: **

*Let z _{1} = a + ib, z_{2} = c + id *

*Re z _{1} = a, Re z_{2} = c, Im z_{1} = b, Im z_{2} = d ….. (1) *

*z _{1}z_{2} = (a + ib) (c + id)*

* = ac + iad + ibc + bd i ^{2}*

* = ac + iad + ibc + bd (-1)*

* = ac + iad + ibc **-** bd *

* = ac **-** bd + i(ad + bc) *

*Comparing real and imaginary part we obtain,*

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

*-*

*bd, Im(z*

_{1}z_{2}) = ad + bc*Now we take real part *

⇒ *Re(z _{1}z_{2}) = ac *

*-*

*bd*

⇒ *Re(z _{1}z_{2}) = Re z_{1} Re z_{2} – Im z_{1} Im z_{2 } [using (1) ] *

**Hence, proved **

Multiplying numerator and denominator by 28 + 10i

On multiplying numerator and denominator by (2 – *i*), we get

On comparing real and imaginary part we obtain

**Q14. ****Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.**

**Solution: **

Let z = (x – iy) (3 + 5i)

= 3x + 5xi - 3yi - 5yi^{2}

= 3x + 5xi - 3yi + 5y

= 3x + 5y + 5xi - 3yi

= (3x + 5y) + (5x - 3y)i

Comparing both sides and equating real and imaginary parts, we get

3x + 5y = – 6 …… (1)

5x - 3y = 24 …… (2)

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them,

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

⇒* x ^{3 }+ 3 . x^{2 }. iy + 3 . x . (iy)*

^{2}*+ (iy)*

^{3}=*u + iv*

⇒* x ^{3 }+ 3x^{2 }y i + 3 x y^{2} i*

^{2}*+ y*

^{3}i^{3}=*u + iv*

⇒* x ^{3 }+ 3x^{2 }y i - 3 x y^{2} *

*- i y*

^{3}=*u + iv*

⇒* x ^{3 }*

*-*

*3 x y*

^{2}+ 3x^{2 }y i*- iy*

^{3}=*u + iv*

⇒*( x ^{3 }*

*-*

*3 x y*

^{2}) +*i*

*(3x*

^{2 }y*-*

*y*

^{3}) =*u + iv*

On equating both sides real and imaginary parts, we get;

u = *x ^{3 }*

*-*

*3 x y*

^{2}, …… (1)*v = 3x ^{2 }y *

*-*

*y*

^{3}, ……(2)

**Q18. ****Find the number of non-zero integral solutions of the equation |1 ****-** *i***| ^{x} = 2^{x}**

^{}

Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0.

**Q19. ****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}**

**Solution : **

(a + ib) (c + id) (e + if) (g + ih) = A + iB ……….. (1) given

Replacing i by (-i) we have

(a - ib) (c - id) (e - if) (g - ih) = A - iB ……….. (2)

Multiplying (1) and (2)

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

(*a*^{2} + *b*^{2}) (*c*^{2} + *d*^{2}) (*e*^{2} + *f *^{2}) (*g*^{2} + *h*^{2}) = A^{2} + B^{2 } [ ∵ (x + iy) (x - iy) = x^{2} + y^{2}]

Hence, proved

Or m = 4k

Hence, least positive integer is 1.

Therefore, the least positive integral value of *m *is 4 × 1 = 4

##### Other Pages of this Chapter: 5. Complex Numbers and Quadratic Equations

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