AHL 1.12 Question Bank

Solution:

Step 1: Use FOIL method

\((3 + 2i)(1 – 4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i)\)

Step 2: Simplify each term

\(= 3 – 12i + 2i – 8i^2\)

Step 3: Substitute \(i^2 = -1\)

\(= 3 – 12i + 2i – 8(-1) = 3 – 10i + 8 = 11 – 10i\)

✅ Answer: \(11 – 10i\)

Solution:

Step 1: Find complex conjugate

For \(z = -3 + 7i\), the conjugate is \(\overline{z} = -3 – 7i\)

Step 2: Calculate \(z \cdot \overline{z}\)

\(z \cdot \overline{z} = (-3 + 7i)(-3 – 7i)\)

\(= (-3)^2 – (7i)^2 = 9 – 49i^2 = 9 – 49(-1) = 9 + 49 = 58\)

✅ Answer: \(\overline{z} = -3 – 7i\), \(z \cdot \overline{z} = 58\)

Solution:

Step 1: Multiply by conjugate of denominator

\(\frac{2 + i}{3 – 2i} \times \frac{3 + 2i}{3 + 2i}\)

Step 2: Calculate numerator

\((2 + i)(3 + 2i) = 6 + 4i + 3i + 2i^2 = 6 + 7i – 2 = 4 + 7i\)

Step 3: Calculate denominator

\((3 – 2i)(3 + 2i) = 9 – 4i^2 = 9 + 4 = 13\)

Step 4: Final result

\(\frac{2 + i}{3 – 2i} = \frac{4 + 7i}{13} = \frac{4}{13} + \frac{7}{13}i\)

✅ Answer: \(\frac{4}{13} + \frac{7}{13}i\)

Solution:

Step 1: Use quadratic formula

For \(az^2 + bz + c = 0\): \(z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Here: \(a = 1\), \(b = -4\), \(c = 13\)

Step 2: Calculate discriminant

\(\Delta = b^2 – 4ac = (-4)^2 – 4(1)(13) = 16 – 52 = -36\)

Step 3: Handle negative discriminant

\(\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i\)

Step 4: Find solutions

\(z = \frac{4 \pm 6i}{2} = 2 \pm 3i\)

✅ Answer: \(z = 2 + 3i\) or \(z = 2 – 3i\)

Solution:

Step 1: Calculate \(z_1^2\)

\(z_1^2 = (1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2\)

\(= 1 + 4i + 4i^2 = 1 + 4i – 4 = -3 + 4i\)

Step 2: Calculate \(2z_2\)

\(2z_2 = 2(3 – i) = 6 – 2i\)

Step 3: Calculate \(z_1^2 – 2z_2\)

\(z_1^2 – 2z_2 = (-3 + 4i) – (6 – 2i) = -3 + 4i – 6 + 2i = -9 + 6i\)

✅ Answer: \(-9 + 6i\)

Solution:

Step 1: Rationalize the denominator

\(\frac{1}{2 – 3i} \times \frac{2 + 3i}{2 + 3i}\)

Step 2: Calculate numerator and denominator

Numerator: \(1 \times (2 + 3i) = 2 + 3i\)

Denominator: \((2 – 3i)(2 + 3i) = 4 – 9i^2 = 4 + 9 = 13\)

Step 3: Express in standard form

\(\frac{1}{2 – 3i} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i\)

✅ Answer: Real part = \(\frac{2}{13}\), Imaginary part = \(\frac{3}{13}\)

Solution:

Step 1: Calculate modulus

\(|z| = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4\)

Step 2: Determine quadrant

Real part: \(-2\sqrt{3} < 0\), Imaginary part: \(2 > 0\)

Therefore, \(z\) is in the second quadrant

Step 3: Find reference angle

\(\tan \theta_{ref} = \frac{|b|}{|a|} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\)

Therefore: \(\theta_{ref} = \frac{\pi}{6}\)

Step 4: Calculate argument

In second quadrant: \(\arg(z) = \pi – \frac{\pi}{6} = \frac{5\pi}{6}\)

✅ Answer: \(|z| = 4\), \(\arg(z) = \frac{5\pi}{6}\)

Solution:

Step 1: Calculate \(w^2\)

\(w^2 = (1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i – 1 = 2i\)

Step 2: Calculate \(w^4 = (w^2)^2\)

\(w^4 = (2i)^2 = 4i^2 = 4(-1) = -4\)

Step 3: Express in standard form

\(w^4 = -4 = -4 + 0i\)

✅ Answer: \(w^4 = -4 + 0i\)

Solution:

Step 1: Interpret geometric meaning

Let \(z = x + yi\) where \(x, y \in \mathbb{R}\)

\(|z – 2i| = |(x + yi) – 2i| = |x + (y – 2)i|\)

Step 2: Apply modulus formula

\(|x + (y – 2)i| = \sqrt{x^2 + (y – 2)^2} = 3\)

Step 3: Square both sides

\(x^2 + (y – 2)^2 = 9\)

Step 4: Identify geometric shape

This is the equation of a circle with center (0, 2) and radius 3

✅ Answer: Circle with center at (0, 2) and radius 3

Solution:

Step 1: Express \(-8i\) in polar form

\(|-8i| = 8\), \(\arg(-8i) = -\frac{\pi}{2}\) (or \(\frac{3\pi}{2}\))

\(-8i = 8e^{-i\pi/2}\)

Step 2: Find cube roots

If \(z^3 = 8e^{-i\pi/2}\), then \(z = 2e^{i(-\pi/6 + 2\pi k/3)}\) for \(k = 0, 1, 2\)

Step 3: Calculate each root

For \(k = 0\): \(z_1 = 2e^{-i\pi/6} = 2(\cos(-\pi/6) + i\sin(-\pi/6)) = 2(\frac{\sqrt{3}}{2} – \frac{i}{2}) = \sqrt{3} – i\)

For \(k = 1\): \(z_2 = 2e^{i\pi/2} = 2i\)

For \(k = 2\): \(z_3 = 2e^{i7\pi/6} = 2(-\frac{\sqrt{3}}{2} – \frac{i}{2}) = -\sqrt{3} – i\)

✅ Answer: \(z = \sqrt{3} – i\), \(z = 2i\), \(z = -\sqrt{3} – i\)

Complete solution:

(a) Product and quotient:

Product: \(z_1 \cdot z_2 = (3 + 4i)(1 – 2i) = 3 – 6i + 4i – 8i^2 = 3 – 2i + 8 = 11 – 2i\)

Quotient: \(\frac{z_1}{z_2} = \frac{3 + 4i}{1 – 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(3 + 4i)(1 + 2i)}{5} = \frac{3 + 6i + 4i + 8i^2}{5} = \frac{-5 + 10i}{5} = -1 + 2i\)

(b) Moduli and verification:

\(|z_1| = \sqrt{3^2 + 4^2} = 5\), \(|z_2| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\)

\(|z_1/z_2| = |-1 + 2i| = \sqrt{1 + 4} = \sqrt{5}\)

\(|z_1|/|z_2| = 5/\sqrt{5} = \sqrt{5}\) ✓

(c) Geometric relationship:

\(z_1 – z_2 = (3 + 4i) – (1 – 2i) = 2 + 6i\)

Points: \(z_1\) at (3, 4), \(z_2\) at (1, -2), \(z_1 – z_2\) at (2, 6)

Geometrically, \(z_1 – z_2\) represents the vector from \(z_2\) to \(z_1\)

✅ Final Answers:
(a) \(z_1 \cdot z_2 = 11 – 2i\), \(z_1/z_2 = -1 + 2i\)
(b) \(|z_1| = 5\), \(|z_2| = \sqrt{5}\), verified
(c) Vector relationship in complex plane

✅ Complete solution showing \(w\) is a primitive 6th root of unity with detailed geometric interpretation

✅ Complete analysis of complex polynomial roots and their geometric properties

✅ Complete investigation of rotations, reflections, and scaling using complex multiplication