Complex Numbers Quiz (50 Questions)

1. The conjugate of the complex number \( z = 3 + 4i \) is:
(KVS PGT 2018)

a) 3 – 4i b) 3 + 4i c) -3 + 4i d) -3 – 4i

Explanation: The conjugate of \( a + bi \) is \( a – bi \). For \( z = 3 + 4i \), conjugate is \( 3 – 4i \).
Correct Answer: a) 3 – 4i

2. The modulus of \( z = -2 + 3i \) is:
(NDA 2019)

a) 2 b) 3 c) \(\sqrt{13}\) d) \(\sqrt{5}\)

Explanation: Modulus of \( z = a + bi \) is \( \sqrt{a^2 + b^2} \). For \( z = -2 + 3i \), modulus = \( \sqrt{(-2)^2 + 3^2} = \sqrt{13} \).
Correct Answer: c) \(\sqrt{13}\)

3. If \( z = 1 + i \), then \( z^2 \) equals:
(UP TGT 2020)

a) 1 + i b) 2i c) -2 d) 1 – i

Explanation: \( z^2 = (1 + i)(1 + i) = 1 + 2i + i^2 = 1 + 2i – 1 = 2i \).
Correct Answer: b) 2i

4. The argument of \( z = -1 – i \) is:
(KVS PGT 2021)

a) \(\pi/4\) b) \(-\pi/4\) c) \(-3\pi/4\) d) \(3\pi/4\)

Explanation: Argument = \( \tan^{-1}(-1/-1) = \tan^{-1}(1) \). Third quadrant, so argument = \( \pi + \pi/4 = -3\pi/4 \).
Correct Answer: c) \(-3\pi/4\)

5. If \( z_1 = 2 + 3i \), \( z_2 = 1 – i \), then \( z_1 + z_2 \):
(NDA 2022)

a) 1 + 2i b) 3 + 2i c) 3 + 4i d) 2 + 4i

Explanation: \( z_1 + z_2 = (2 + 3i) + (1 – i) = 3 + 2i \).
Correct Answer: b) 3 + 2i

6. The product \( z_1 z_2 \) where \( z_1 = 1 + i \), \( z_2 = 1 – i \):
(UP PGT 2019)

a) 2i b) 1 + i c) 2 d) -2

Explanation: \( (1 + i)(1 – i) = 1 – i^2 = 1 – (-1) = 2 \).
Correct Answer: c) 2

7. The value of \( i^{10} \):
(LT Grade 2020)

a) i b) -i c) -1 d) 1

Explanation: \( i^{10} = (i^4)^2 \cdot i^2 = 1^2 \cdot (-1) = -1 \).
Correct Answer: c) -1

8. The real part of \( \frac{1}{1 + i} \):
(PCS 2017)

a) 1 b) \(\frac{1}{2}\) c) 0 d) -1

Explanation: \( \frac{1}{1 + i} \cdot \frac{1 – i}{1 – i} = \frac{1 – i}{2} = \frac{1}{2} – \frac{i}{2} \). Real part = \( \frac{1}{2} \).
Correct Answer: b) \(\frac{1}{2}\)

9. The imaginary part of \( (2 – 3i)^2 \):
(KVS TGT 2018)

a) -12 b) 12 c) -12i d) 12i

Explanation: \( (2 – 3i)^2 = 4 – 12i + 9i^2 = 4 – 12i – 9 = -5 – 12i \). Imaginary part = -12.
Correct Answer: a) -12

10. The polar form of \( z = 1 + i \):
(NDA 2020)

a) \(\sqrt{2} (\cos \pi/2 + i \sin \pi/2)\) b) \(\sqrt{2} (\cos \pi/4 + i \sin \pi/4)\) c) \(2 (\cos \pi/4 + i \sin \pi/4)\) d) \(\sqrt{2} (\cos \pi + i \sin \pi)\)

Explanation: Modulus = \( \sqrt{1^2 + 1^2} = \sqrt{2} \), argument = \( \tan^{-1}(1) = \pi/4 \). Polar form = \( \sqrt{2} (\cos \pi/4 + i \sin \pi/4) \).
Correct Answer: b) \(\sqrt{2} (\cos \pi/4 + i \sin \pi/4)\)

11. The value of \( \frac{i}{1 – i} \):
(PCS 2023)

a) \(1 + i\) b) \(\frac{1}{2} – \frac{i}{2}\) c) \(\frac{1}{2} + \frac{i}{2}\) d) \(1 – i\)

Explanation: \( \frac{i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{i + i^2}{1 – i^2} = \frac{i – 1}{2} = \frac{1}{2} + \frac{i}{2} \).
Correct Answer: c) \(\frac{1}{2} + \frac{i}{2}\)

12. If \( z = 3 – 4i \), then \( |z|^2 \):
(KVS TGT 2024)

a) 7 b) 9 c) 25 d) 16

Explanation: \( |z|^2 = 3^2 + (-4)^2 = 9 + 16 = 25 \).
Correct Answer: c) 25

13. The value of \( (1 + i)^3 \):
(UP TGT 2021)

a) 1 + i b) -2 + 2i c) 2 – 2i d) -1 – i

Explanation: \( (1 + i)^2 = 2i \), so \( (1 + i)^3 = 2i (1 + i) = 2i + 2i^2 = 2i – 2 = -2 + 2i \).
Correct Answer: b) -2 + 2i

14. The modulus of \( z = -5 + 12i \):
(NDA 2021)

a) 5 b) 12 c) 13 d) \(\sqrt{119}\)

Explanation: Modulus = \( \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).
Correct Answer: c) 13

15. The conjugate of \( z = 7 – 2i \):
(KVS PGT 2020)

a) -7 + 2i b) 7 + 2i c) 7 – 2i d) -7 – 2i

Explanation: Conjugate of \( 7 – 2i \) is \( 7 + 2i \).
Correct Answer: b) 7 + 2i

16. The value of \( \frac{1 + i}{1 – i} \):
(UP PGT 2022)

a) -i b) 1 c) i d) -1

Explanation: \( \frac{1 + i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1 – i^2} = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i \).
Correct Answer: c) i

17. The argument of \( z = 1 – i \):
(NDA 2023)

a) \(\pi/4\) b) \(-\pi/4\) c) \(\pi/2\) d) \(-\pi/2\)

Explanation: Argument = \( \tan^{-1}(-1/1) = -\pi/4 \). Fourth quadrant.
Correct Answer: b) \(-\pi/4\)

18. If \( z_1 = 3 + 2i \), \( z_2 = 2 – i \), then \( z_1 – z_2 \):
(KVS TGT 2021)

a) 1 + i b) 1 + 3i c) 5 + i d) 1 – i

Explanation: \( z_1 – z_2 = (3 + 2i) – (2 – i) = 1 + 3i \).
Correct Answer: b) 1 + 3i

19. The value of \( i^{25} \):
(LT Grade 2021)

a) i b) -i c) 1 d) -1

Explanation: \( i^{25} = i^{24+1} = (i^4)^6 \cdot i = 1^6 \cdot i = i \).
Correct Answer: a) i

20. The polar form of \( z = -1 + i \):
(UP PGT 2020)

a) \(\sqrt{2} (\cos \pi/4 + i \sin \pi/4)\) b) \(\sqrt{2} (\cos 3\pi/4 + i \sin 3\pi/4)\) c) \(\sqrt{2} (\cos \pi/2 + i \sin \pi/2)\) d) \(2 (\cos \pi/4 + i \sin \pi/4)\)

Explanation: Modulus = \( \sqrt{(-1)^2 + 1^2} = \sqrt{2} \), argument = \( \tan^{-1}(1/-1) = 3\pi/4 \). Polar form = \( \sqrt{2} (\cos 3\pi/4 + i \sin 3\pi/4) \).
Correct Answer: b) \(\sqrt{2} (\cos 3\pi/4 + i \sin 3\pi/4)\)

21. The real part of \( \frac{2 + i}{1 – i} \):
(KVS PGT 2023)

a) 1 b) \(\frac{1}{2}\) c) 2 d) 0

Explanation: \( \frac{2 + i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(2 + i)(1 + i)}{2} = \frac{1 + 3i}{2} = \frac{1}{2} + \frac{3i}{2} \). Real part = \( \frac{1}{2} \).
Correct Answer: b) \(\frac{1}{2}\)

22. The value of \( (1 – i)^4 \):
(NDA 2020)

a) -4 b) 4i c) 4 d) -4i

Explanation: \( (1 – i)^2 = -2i \), so \( (1 – i)^4 = (-2i)^2 = 4 \).
Correct Answer: c) 4

23. The modulus of \( z = 4 – 3i \):
(UP TGT 2022)

a) 3 b) 4 c) 5 d) 7

Explanation: Modulus = \( \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5 \).
Correct Answer: c) 5

24. The argument of \( z = -2 \):
(KVS PGT 2021)

a) 0 b) \(\pi\) c) \(\pi/2\) d) \(-\pi/2\)

Explanation: For \( z = -2 \), argument = \( \tan^{-1}(0/-2) = \pi \).
Correct Answer: b) \(\pi\)

25. The value of \( z \bar{z} \) for \( z = 2 + 3i \):
(LT Grade 2020)

a) 5 b) 7 c) 13 d) 4

Explanation: \( z \bar{z} = (2 + 3i)(2 – 3i) = 4 + 9 = 13 \).
Correct Answer: c) 13

26. The imaginary part of \( \frac{1 + 2i}{1 – i} \):
(PCS 2020)

a) 1 b) 2 c) \(\frac{3}{2}\) d) 0

Explanation: \( \frac{1 + 2i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + 2i)(1 + i)}{2} = \frac{-1 + 3i}{2} \). Imaginary part = \( \frac{3}{2} \).
Correct Answer: c) \(\frac{3}{2}\)

27. The value of \( (2 + i)^2 – (2 – i)^2 \):
(NDA 2022)

a) 4i b) 8i c) 4 d) -8i

Explanation: \( (2 + i)^2 = 3 + 4i \), \( (2 – i)^2 = 3 – 4i \). Subtract: \( (3 + 4i) – (3 – 4i) = 8i \).
Correct Answer: b) 8i

28. The conjugate of \( z = \sqrt{2} (\cos \pi/6 + i \sin \pi/6) \):
(UP PGT 2021)

a) \(\sqrt{2} (\cos \pi/6 – i \sin \pi/6)\) b) \(\sqrt{2} (\cos (-\pi/6) + i \sin (-\pi/6))\) c) \(\sqrt{2} (\cos \pi/6 + i \sin \pi/6)\) d) \(\sqrt{2} (\cos \pi/3 – i \sin \pi/3)\)

Explanation: Conjugate of \( r (\cos \theta + i \sin \theta) \) is \( r (\cos \theta – i \sin \theta) \). Thus, conjugate is \( \sqrt{2} (\cos \pi/6 – i \sin \pi/6) \).
Correct Answer: a) \(\sqrt{2} (\cos \pi/6 – i \sin \pi/6)\)

29. The value of \( i^{100} \):
(KVS TGT 2020)

a) i b) -i c) 1 d) -1

Explanation: \( i^{100} = (i^4)^{25} = 1^{25} = 1 \).
Correct Answer: c) 1

30. The modulus of \( z = -3 – 4i \):
(NDA 2022)

a) 3 b) 4 c) 5 d) 7

Explanation: Modulus = \( \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5 \).
Correct Answer: c) 5

31. The real part of \( (3 – 2i)^2 \):
(UP TGT 2020)

a) 13 b) 5 c) -5 d) 0

Explanation: \( (3 – 2i)^2 = 9 – 12i + 4i^2 = 5 – 12i \). Real part = 5.
Correct Answer: b) 5

32. The argument of \( z = i \):
(KVS PGT 2022)

a) 0 b) \(\pi/2\) c) \(\pi\) d) \(-\pi/2\)

Explanation: For \( z = i \), argument = \( \tan^{-1}(1/0) = \pi/2 \).
Correct Answer: b) \(\pi/2\)

33. The value of \( \frac{1 – i}{1 + i} \):
(PCS 2021)

a) 1 b) -1 c) -i d) i

Explanation: \( \frac{1 – i}{1 + i} \cdot \frac{1 – i}{1 – i} = \frac{(1 – i)^2}{1 – i^2} = \frac{-2i}{2} = -i \).
Correct Answer: c) -i

34. The polar form of \( z = -i \):
(NDA 2020)

a) \(\cos \pi/2 + i \sin \pi/2\) b) \(\cos (-\pi/2) + i \sin (-\pi/2)\) c) \(\cos \pi + i \sin \pi\) d) \(\cos 0 + i \sin 0\)

Explanation: Modulus = 1, argument = \( \tan^{-1}(-1/0) = -\pi/2 \). Polar form = \( \cos (-\pi/2) + i \sin (-\pi/2) \).
Correct Answer: b) \(\cos (-\pi/2) + i \sin (-\pi/2)\)

35. The value of \( z_1 z_2 \) where \( z_1 = 2 – i \), \( z_2 = 3 + i \):
(UP PGT 2020)

a) 5 + i b) 7 – i c) 7 + i d) 5 – i

Explanation: \( (2 – i)(3 + i) = 6 + 3i – 2i – i^2 = 7 – i \).
Correct Answer: b) 7 – i

36. The imaginary part of \( \frac{3 – i}{2 + i} \):
(KVS TGT 2021)

a) 1 b) -1 c) \(-\frac{1}{5}\) d) 0

Explanation: \( \frac{3 – i}{2 + i} \cdot \frac{2 – i}{2 – i} = \frac{(3 – i)(2 – i)}{5} = \frac{5 – 5i}{5} = 1 – i \). Imaginary part = -1.
Correct Answer: b) -1

37. The value of \( (1 + i)^8 \):
(NDA 2021)

a) 16i b) -16 c) 16 d) -16i

Explanation: \( (1 + i)^2 = 2i \), so \( (1 + i)^8 = (2i)^4 = 16i^4 = 16 \).
Correct Answer: c) 16

38. The modulus of \( z = 1 – i\sqrt{3} \):
(UP PGT 2021)

a) 1 b) \(\sqrt{3}\) c) 2 d) 4

Explanation: Modulus = \( \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2 \).
Correct Answer: c) 2

39. The argument of \( z = 1 – i\sqrt{3} \):
(KVS PGT 2020)

a) \(\pi/3\) b) \(-\pi/3\) c) \(\pi/6\) d) \(-\pi/6\)

Explanation: Argument = \( \tan^{-1}(-\sqrt{3}/1) = -\pi/3 \).
Correct Answer: b) \(-\pi/3\)

40. The value of \( z_1 / z_2 \) where \( z_1 = 4 + 3i \), \( z_2 = 2 – i \):
(NDA 2020)

a) 1 + i b) 1 + 2i c) 2 – i d) 1 – i

Explanation: \( \frac{4 + 3i}{2 – i} \cdot \frac{2 + i}{2 + i} = \frac{(4 + 3i)(2 + i)}{5} = \frac{5 + 10i}{5} = 1 + 2i \).
Correct Answer: b) 1 + 2i

41. The real part of \( (1 – i)^3 \):
(UP TGT 2021)

a) 2 b) -2 c) 0 d) 1

Explanation: \( (1 – i)^2 = -2i \), so \( (1 – i)^3 = -2i (1 – i) = -2i + 2i^2 = -2 – 2i \). Real part = -2.
Correct Answer: b) -2

42. The modulus of \( z = 2 + i\sqrt{2} \):
(KVS PGT 2021)

a) 2 b) \(\sqrt{2}\) c) \(\sqrt{6}\) d) 4

Explanation: Modulus = \( \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{4 + 2} = \sqrt{6} \).
Correct Answer: c) \(\sqrt{6}\)

43. The argument of \( z = \sqrt{3} + i \):
(NDA 2021)

a) \(\pi/3\) b) \(\pi/6\) c) \(\pi/4\) d) \(\pi/2\)

Explanation: Argument = \( \tan^{-1}(1/\sqrt{3}) = \pi/6 \).
Correct Answer: b) \(\pi/6\)

44. The value of \( (2 + i)^3 \):
(UP PGT 2020)

a) 2 + 11i b) 2 – 11i c) 8 + i d) 8 – i

Explanation: \( (2 + i)^2 = 3 + 4i \), so \( (2 + i)^3 = (3 + 4i)(2 + i) = 2 – 11i \).
Correct Answer: b) 2 – 11i

45. The real part of \( \frac{4 + i}{2 – i} \):
(KVS TGT 2020)

a) 1

46. The value of \( (1 + i)^4 \):
(LT Grade 2021)

a) 4i b) -4i c) -4 d) 4

Explanation: \( (1 + i)^2 = 2i \), so \( (1 + i)^4 = (2i)^2 = -4 \).
Correct Answer: c) -4

47. The modulus of \( z = 5 + 12i \):
(PCS 2022)

a) 5 b) 12 c) 13 d) 17

Explanation: Modulus = \( \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 \).
Correct Answer: c) 13

48. The argument of \( z = -1 + i\sqrt{3} \):
(NDA 2022)

a) \(\pi/3\) b) \(2\pi/3\) c) \(-\pi/3\) d) \(\pi/2\)

Explanation: Argument = \( \tan^{-1}(\sqrt{3}/-1) = 2\pi/3 \) (second quadrant).
Correct Answer: b) \(2\pi/3\)

49. The value of \( \frac{1}{i} \):
(KVS PGT 2022)

a) i b) -i c) 1 d) -1

Explanation: \( \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = -i \).
Correct Answer: b) -i

50. The value of \( (1 + \omega + \omega^2)^3 \), where \( \omega \) is cube root of unity:
(UP TGT 2022)

a) 1 b) -1 c) 0 d) 3

Explanation: \( 1 + \omega + \omega^2 = 0 \), so \( (1 + \omega + \omega^2)^3 = 0 \).
Correct Answer: c) 0

Complex Numbers Quiz (50 Questions)

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Chandra Shekhar

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Complex Numbers Quiz (50 Questions)

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Chandra Shekhar

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