# Class XI Chapter 5 Complex Numbers and Quadratic Equations Maths. Exercise 5.1. Page 1 of 34

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1 Question 1: Exercise 5.1 Express the given complex number in the form a + ib: Question 2: Express the given complex number in the form a + ib: i 9 + i 19 Question 3: Express the given complex number in the form a + ib: i 39 Page 1 of 34

2 Question 4: Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7) Question 5: Express the given complex number in the form a + ib: (1 i) ( 1 + i6) Question 6: Express the given complex number in the form a + ib: Question 7: Express the given complex number in the form a + ib: Page 2 of 34

3 Question 8: Express the given complex number in the form a + ib: (1 i) 4 Question 9: Express the given complex number in the form a + ib: Page 3 of 34

4 Question 10: Express the given complex number in the form a + ib: Question 11: Find the multiplicative inverse of the complex number 4 3i Page 4 of 34

5 Let z = 4 3i Then, = 4 + 3i and Therefore, the multiplicative inverse of 4 3i is given by Question 12: Find the multiplicative inverse of the complex number Let z = Therefore, the multiplicative inverse of is given by Question 13: Find the multiplicative inverse of the complex number i Let z = i Therefore, the multiplicative inverse of i is given by Question 14: Page 5 of 34

6 Express the following expression in the form of a + ib. Page 6 of 34

7 Exercise 5.2 Question 1: Find the modulus and the argument of the complex number On squaring and adding, we obtain Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant, Thus, the modulus and argument of the complex number are 2 and respectively. Question 2: Page 7 of 34

8 Find the modulus and the argument of the complex number On squaring and adding, we obtain Thus, the modulus and argument of the complex number are 2 and respectively. Question 3: Convert the given complex number in polar form: 1 i 1 i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain Page 8 of 34

9 the required polar form. This is Question 4: Convert the given complex number in polar form: 1 + i 1 + i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain It can be written, Page 9 of 34

10 This is the required polar form. Question 5: Convert the given complex number in polar form: 1 i 1 i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain required polar form. This is the Question 6: Convert the given complex number in polar form: 3 3 Let r cos θ = 3 and r sin θ = 0 On squaring and adding, we obtain Page 10 of 34

11 This is the required polar form. Question 7: Convert the given complex number in polar form: Let r cos θ = and r sin θ = 1 On squaring and adding, we obtain Page 11 of 34

12 This is the required polar form. Question 8: Convert the given complex number in polar form: i i Let r cosθ = 0 and r sin θ = 1 On squaring and adding, we obtain This is the required polar form. Page 12 of 34

13 Exercise 5.3 Question 1: Solve the equation x = 0 The given quadratic equation is x = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 0, and c = 3 Therefore, the discriminant of the given equation is D = b 2 4ac = = 12 Therefore, the required solutions are Question 2: Solve the equation 2x 2 + x + 1 = 0 The given quadratic equation is 2x 2 + x + 1 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 2, b = 1, and c = 1 Therefore, the discriminant of the given equation is D = b 2 4ac = = 1 8 = 7 Therefore, the required solutions are Page 13 of 34

14 Question 3: Solve the equation x 2 + 3x + 9 = 0 The given quadratic equation is x 2 + 3x + 9 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 3, and c = 9 Therefore, the discriminant of the given equation is D = b 2 4ac = = 9 36 = 27 Therefore, the required solutions are Question 4: Solve the equation x 2 + x 2 = 0 The given quadratic equation is x 2 + x 2 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 1, and c = 2 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 1) ( 2) = 1 8 = 7 Therefore, the required solutions are Question 5: Solve the equation x 2 + 3x + 5 = 0 The given quadratic equation is x 2 + 3x + 5 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain Page 14 of 34

15 a = 1, b = 3, and c = 5 Therefore, the discriminant of the given equation is D = b 2 4ac = =9 20 = 11 Therefore, the required solutions are Question 6: Solve the equation x 2 x + 2 = 0 The given quadratic equation is x 2 x + 2 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 1, and c = 2 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 1) = 1 8 = 7 Therefore, the required solutions are Question 7: Solve the equation The given quadratic equation is On comparing the given equation with ax 2 + bx + c = 0, we obtain a =, b = 1, and c = Therefore, the discriminant of the given equation is D = b 2 4ac = 1 2 = 1 8 = 7 Therefore, the required solutions are Page 15 of 34

16 Question 8: Solve the equation The given quadratic equation is On comparing the given equation with ax 2 + bx + c = 0, we obtain a =, b =, and c = Therefore, the discriminant of the given equation is D = b 2 4ac = Therefore, the required solutions are Question 9: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a =, b =, and c = 1 Therefore, the required solutions are Page 16 of 34

17 Question 10: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a =, b = 1, and c = Therefore, the required solutions are Page 17 of 34

18 NCERT Miscellaneous Solutions Question 1: Evaluate: Page 18 of 34

19 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 Page 19 of 34

20 Question 3: Reduce to the standard form. Page 20 of 34

21 Question 4: If x iy = prove that. Page 21 of 34

22 Question 5: Convert the following in the polar form: (i), (ii) (i) Here, Let r cos θ = 1 and r sin θ = 1 Page 22 of 34

23 On squaring and adding, we obtain r 2 (cos 2 θ + sin 2 θ) = r 2 (cos 2 θ + sin 2 θ) = 2 r 2 = 2 [cos 2 θ + sin 2 θ = 1] z = r cos θ + i r sin θ This is the required polar form. (ii) Here, Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain r 2 (cos 2 θ + sin 2 θ) = r 2 (cos 2 θ + sin 2 θ) = 2 r 2 = 2 [cos 2 θ + sin 2 θ = 1] Page 23 of 34

24 z = r cos θ + i r sin θ This is the required polar form. Question 6: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a = 9, b = 12, and c = 20 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 12) = = 576 Therefore, the required solutions are Question 7: Solve the equation Page 24 of 34

25 The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a = 2, b = 4, and c = 3 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 4) = = 8 Therefore, the required solutions are Question 8: Solve the equation 27x 2 10x + 1 = 0 The given quadratic equation is 27x 2 10x + 1 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 27, b = 10, and c = 1 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 10) = = 8 Therefore, the required solutions are Question 9: Solve the equation 21x 2 28x + 10 = 0 The given quadratic equation is 21x 2 28x + 10 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain Page 25 of 34

26 a = 21, b = 28, and c = 10 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 28) = = 56 Therefore, the required solutions are Question 10: If find. Question 11: Page 26 of 34

27 If a + ib =, prove that a 2 + b 2 = On comparing real and imaginary parts, we obtain Hence, proved. Question 12: Let. Find Page 27 of 34

28 (i), (ii) (i) On multiplying numerator and denominator by (2 i), we obtain On comparing real parts, we obtain (ii) On comparing imaginary parts, we obtain Question 13: Find the modulus and argument of the complex number. Let, then Page 28 of 34

29 On squaring and adding, we obtain Therefore, the modulus and argument of the given complex number are respectively. Question 14: Find the real numbers x and y if (x iy) (3 + 5i) is the conjugate of 6 24i. Let It is given that, Page 29 of 34

30 Equating real and imaginary parts, we obtain Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain Putting the value of x in equation (i), we obtain Thus, the values of x and y are 3 and 3 respectively. Question 15: Find the modulus of. Question 16: Page 30 of 34

31 If (x + iy) 3 = u + iv, then show that. On equating real and imaginary parts, we obtain Hence, proved. Question 17: If α and β are different complex numbers with = 1, then find. Let α = a + ib and β = x + iy It is given that, Page 31 of 34

32 Question 18: Find the number of non-zero integral solutions of the equation. Page 32 of 34

33 Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0. Question 19: 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. 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 20: If, then find the least positive integral value of m. Page 33 of 34

34 Therefore, the least positive integer is 1. Thus, the least positive integral value of m is 4 (= 4 1). Page 34 of 34

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