Complex Numbers: Operations, Polar Form, and De Moivre's
Theorem - 100 MCQ Problems with Solutions
1. What is the result of (3 + 4i) + (2 - 5i)?
A) 5 - i
B) 5 - 9i
C) 5 + 9i
D) 5 + i
Answer: A) 5 - i
Solution:
(3 + 4i) + (2 - 5i) = (3 + 2) + (4i - 5i) = 5 - i
2. Simplify (2 + 3i)(4 - i).
A) 11 + 10i
B) 11 - 10i
C) 8 + 11i
D) 8 - 11i
Answer: A) 11 + 10i
Solution:
(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2
= 8 + 10i + 3 (since i^2 = -1)
= 11 + 10i
3. Find the modulus of 3 - 4i.
A) 5
B) 7
C) 1
D) 25
Answer: A) 5
Solution:
|3 - 4i| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5
4. What is the conjugate of 2 + 5i?
A) 2 - 5i
B) -2 + 5i
C) -2 - 5i
D) 5 - 2i
Answer: A) 2 - 5i
Solution:
The conjugate of a + bi is a - bi.
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
5. Express 1 + i in polar form.
A) √2(cos(π/4) + i sin(π/4))
B) 2(cos(π/4) + i sin(π/4))
C) √2(cos(π/2) + i sin(π/2))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) √2(cos(π/4) + i sin(π/4))
Solution:
r = √(1^2 + 1^2) = √2
θ = tan^(-1)(1/1) = π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(π/4) + i sin(π/4))
6. Calculate (1 + i)^4 using De Moivre's Theorem.
A) -4
B) 4
C) 4i
D) -4i
Answer: A) -4
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^4 = (√2)^4(cos(4π/4) + i sin(4π/4))
= 4(cos(π) + i sin(π))
= 4(-1 + 0i)
= -4
7. What is the result of i^17?
A) i
B) -i
C) 1
D) -1
Answer: B) -i
Solution:
i^4 = 1, so i^16 = 1
i^17 = i^16 * i = 1 * i = i
i * i = -1, so i^17 = -i
8. Find the roots of x^2 + 4x + 13 = 0.
A) -2 + 3i and -2 - 3i
B) 2 + 3i and 2 - 3i
C) -2 + i and -2 - i
D) 2 + i and 2 - i
Answer: A) -2 + 3i and -2 - 3i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 - 52)) / 2
x = (-4 ± √(-36)) / 2
x = (-4 ± 6i) / 2
x = -2 ± 3i
9. What is the polar form of -1?
A) cos(π) + i sin(π)
B) cos(π/2) + i sin(π/2)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π) + i sin(π)
Solution:
r = |-1| = 1
θ = π (angle in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π) + i sin(π)
10. Simplify (1 + i)/(1 - i).
A) -i
B) i
C) 1 + i
D) 1 - i
Answer: B) i
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + 2i - 1)/(1 + 1) = 2i/2 = i
11. What is the result of e^(iπ)?
A) 1
B) -1
C) i
D) -i
Answer: B) -1
Solution:
By Euler's formula: e^(iπ) = cos(π) + i sin(π) = -1 + 0i = -1
12. Find the magnitude of 3 + 4i.
A) 5
B) 7
C) 25
D) 1
Answer: A) 5
Solution:
|3 + 4i| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
13. What is the polar form of i?
A) cos(π/2) + i sin(π/2)
B) cos(π) + i sin(π)
C) cos(3π/2) + i sin(3π/2)
D) cos(2π) + i sin(2π)
Answer: A) cos(π/2) + i sin(π/2)
Solution:
r = |i| = 1
θ = π/2 (90 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(π/2) + i sin(π/2)
14. Calculate (1 + i)^6 using De Moivre's Theorem.
A) -8
B) 8
C) 8i
D) -8i
Answer: B) 8
Solution:
(1 + i) = √2(cos(π/4) + i sin(π/4))
(1 + i)^6 = (√2)^6(cos(6π/4) + i sin(6π/4))
= 8(cos(3π/2) + i sin(3π/2))
= 8(0 - i)
= -8i
15. What is the result of i^28?
A) 1
B) -1
C) i
D) -i
Answer: A) 1
Solution:
i^4 = 1, so i^28 = (i^4)^7 = 1^7 = 1
16. Find the roots of x^2 - 6x + 13 = 0.
A) 3 + 2i and 3 - 2i
B) 3 + i and 3 - i
C) -3 + 2i and -3 - 2i
D) -3 + i and -3 - i
Answer: A) 3 + 2i and 3 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (6 ± √(36 - 52)) / 2
x = (6 ± √(-16)) / 2
x = (6 ± 4i) / 2
x = 3 ± 2i
17. What is the polar form of -i?
A) cos(3π/2) + i sin(3π/2)
B) cos(π/2) + i sin(π/2)
C) cos(π) + i sin(π)
D) cos(2π) + i sin(2π)
Answer: A) cos(3π/2) + i sin(3π/2)
Solution:
r = |-i| = 1
θ = 3π/2 (270 degrees in the complex plane)
Polar form: r(cos(θ) + i sin(θ)) = cos(3π/2) + i sin(3π/2)
18. Simplify (3 - 2i)/(4 + 3i).
A) (18 - 17i)/25
B) (18 + 17i)/25
C) (17 - 18i)/25
D) (17 + 18i)/25
Answer: A) (18 - 17i)/25
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(3 - 2i)/(4 + 3i) * (4 - 3i)/(4 - 3i)
= (12 - 9i - 8i + 6i^2) / (16 + 9)
= (12 - 17i + 6) / 25
= (18 - 17i) / 25
19. What is the result of e^(iπ/2)?
A) 1
B) -1
C) i
D) -i
Answer: C) i
Solution:
By Euler's formula: e^(iπ/2) = cos(π/2) + i sin(π/2) = 0 + i = i
20. Find the magnitude of 1 - i.
A) √2
B) 2
C) 1
D) 0
Answer: A) √2
Solution:
|1 - i| = √(1^2 + (-1)^2) = √(1 + 1) = √2
21. What is the polar form of 1 - i?
A) √2(cos(-π/4) + i sin(-π/4))
B) √2(cos(π/4) + i sin(π/4))
C) √2(cos(3π/4) + i sin(3π/4))
D) √2(cos(5π/4) + i sin(5π/4))
Answer: A) √2(cos(-π/4) + i sin(-π/4))
Solution:
r = |1 - i| = √2
θ = tan^(-1)(-1/1) = -π/4
Polar form: r(cos(θ) + i sin(θ)) = √2(cos(-π/4) + i sin(-π/4))
22. Calculate (1 - i)^8 using De Moivre's Theorem.
A) 16
B) -16
C) 16i
D) -16i
Answer: A) 16
Solution:
(1 - i) = √2(cos(-π/4) + i sin(-π/4))
(1 - i)^8 = (√2)^8(cos(-8π/4) + i sin(-8π/4))
= 16(cos(-2π) + i sin(-2π))
= 16(1 + 0i)
= 16
23. What is the result of i^35?
A) i
B) -i
C) 1
D) -1
Answer: A) i
Solution:
i^4 = 1, so i^32 = 1
i^35 = i^32 * i^3 = 1 * i^3 = -i * i = i
24. Find the roots of x^2 + 2x + 5 = 0.
A) -1 + 2i and -1 - 2i
B) 1 + 2i and 1 - 2i
C) -1 + i and -1 - i
D) 1 + i and 1 - i
Answer: A) -1 + 2i and -1 - 2i
Solution:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
x = (-2 ± 4i) / 2
x = -1 ± 2i
25. What is the polar form of 2 + 2i?
A) 2√2(cos(π/4) + i sin(π/4))
B) 2√2(cos(π/2) + i sin(π/2))
C) 2(cos(π/4) + i sin(π/4))
D) 2(cos(π/2) + i sin(π/2))
Answer: A) 2√2(cos(π/4) + i sin(π/4))
Solution:
r = |2 + 2i| = √(2^2 + 2^2) = √8 = 2√2
θ = tan^(-1)(2/2) = π/4
Polar form: r(cos(θ) + i sin(θ)) = 2√2(cos(π/4) + i sin(π/4))
26. Simplify (2 + 3i)/(1 - i).
A) 5/2 + i/2
B) -5/2 - i/2
C) 5/2 - i/2
D) -5/2 + i/2
Answer: A) 5/2 + i/2
Solution:
Multiply numerator and denominator by the conjugate of the denominator:
(2 + 3i)/(1 - i) * (1 + i)/(1 + i)
= (2 + 2i + 3i - 3) / (1 + 1)
= (-1 + 5i) / 2
= -1/2 + 5i/2
= 5/2 + i/2
27. What is the result of e^(iπ/3)?
A) 1/2 + (√3/2)i
B) √3/2 + (1/2)i
C) -1/2 + (√3/2)i
D) -√3/2 + (1/2)i
Answer: A) 1/2 + (√3/2)i
Solution:
By Euler's formula: e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
28. Find the magnitude of 2 - 3i.
