Question 1: Express (5i) × (-3/5 i) in the form a + ib.

Solution:

Given expression = (5i) × (-3/5 i)

= 5 × (-3/5) × (i × i)

= -3 × i²

Since i² = -1:

= -3 × (-1) = 3

In standard form a + ib, this is written as 3 + 0i.

Question 2: Express i⁹ + i¹⁹ in the form a + ib.

Solution:

Given expression = i⁹ + i¹⁹

Using powers of i:

i⁹ = (i⁴)² × i = (1)² × i = i

i¹⁹ = (i⁴)⁴ × i³ = (1)⁴ × (-i) = -i

Therefore, i⁹ + i¹⁹ = i + (-i) = 0

In standard form a + ib, this is 0 + 0i.

Question 3: Express i^-39 in the form a + ib.

Solution:

Given expression = i^-39 = 1 / i³⁹

Now, i³⁹ = (i⁴)⁹ × i³ = (1)⁹ × (-i) = -i

So, i^-39 = 1 / (-i) = -1/i

Multiply numerator and denominator by i:

= (-1 × i) / (i × i) = -i / i² = -i / (-1) = i

In standard form a + ib, this is 0 + 1i (or 0 + i).

Question 4: Find the multiplicative inverse of the complex number 4 - 3i.

Solution:

Let z = 4 - 3i.

Here, a = 4 and b = -3.

Modulus squared: |z|² = a² + b² = 4² + (-3)² = 16 + 9 = 25.

Conjugate of z: z̄ = 4 + 3i.

Multiplicative inverse z^-1 = z̄ / |z|² = (4 + 3i) / 25 = (4/25) + (3/25)i.

Question 5: Express [(3 + i√5)(3 - i√5)] / [(√3 + √2i) - (√3 - i√2)] in the form a + ib.

Solution:

Numerator = (3 + i√5)(3 - i√5) = 3² - (i√5)² = 9 - i²(5) = 9 - (-1)(5) = 9 + 5 = 14.

Denominator = (√3 + √2i) - (√3 - i√2) = √3 + √2i - √3 + √2i = 2√2i.

Expression = 14 / (2√2i) = 7 / (√2i)

Multiply numerator and denominator by i:

= (7i) / (√2 i²) = (7i) / (-√2) = -7i / √2

Rationalizing: = (-7√2 / 2)i

In standard form: 0 - (7√2 / 2)i.

Question 6: Solve the quadratic equation x² + 3 = 0.

Solution:

Given equation: x² + 3 = 0

x² = -3

x = ±√(-3) = ±√(3 × -1) = ±√3 i

Hence, the solutions are x = √3 i, -√3 i.

Question 7: Solve the quadratic equation 2x² + x + 1 = 0.

Solution:

Comparing with ax² + bx + c = 0, we get a = 2, b = 1, c = 1.

Discriminant D = b² - 4ac = 1² - 4(2)(1) = 1 - 8 = -7.

Using quadratic formula: x = [-b ± √D] / 2a

x = [-1 ± √(-7)] / (2 × 2) = [-1 ± √7 i] / 4

Hence, the solutions are x = (-1 + √7 i) / 4 and x = (-1 - √7 i) / 4.

Question 8: Find the modulus and argument of the complex number z = -1 - i√3.

Solution:

Let z = -1 - i√3 = a + ib. Here a = -1 and b = -√3.

Modulus (r): r = |z| = √(a² + b²) = √[(-1)² + (-√3)²] = √(1 + 3) = √4 = 2.

Argument (θ): Let tan α = |b/a| = |(-√3)/(-1)| = √3 ⇒ α = π/3.

Since both a and b are negative (a < 0, b < 0), z lies in the 3rd quadrant.

Argument θ = -(π - α) = -(π - π/3) = -2π/3.

Hence, Modulus = 2 and Argument = -2π/3 (or 240°).