Free WACE Mathematics Specialist Flashcard Review

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What are Re(z) and Im(z) for z = 4 − 7i?

Re(z) = 4 and Im(z) = −7. The imaginary part is the real coefficient of i, so it does not include i itself.

Express (3 + 2i)(1 − 4i) in Cartesian form.

Expand and use i² = −1: 3 − 12i + 2i − 8i² = 3 − 10i + 8 = 11 − 10i.

How do you write (2 + i)/(1 − 2i) in Cartesian form?

Multiply top and bottom by the conjugate 1 + 2i: (2 + i)(1 + 2i)/(1 + 4) = (2 + 5i − 2)/5 = i. So the quotient is 0 + 1i.

What is z z̄ for z = a + bi, and why is it useful?

z z̄ = a² + b² = |z|², a real number. Multiplying by the conjugate is how you remove i from a denominator and how you prove modulus identities.

If z = 2 − 3i, what is z⁻¹ in Cartesian form?

z⁻¹ = z̄/|z|² = (2 + 3i)/13 = 2/13 + (3/13)i.

What is the principal argument Arg(z) and its range?

Arg(z) is the angle from the positive real axis to the vector for z, chosen in −π < Arg(z) ≤ π. It is undefined for z = 0.

State |z| and Arg(z) for z = −2 − 2i.

|z| = √8 = 2√2. The point is in the third quadrant with reference angle π/4, so Arg(z) = −3π/4.

Write 4 cis(2π/3) in Cartesian form.

4(cos(2π/3) + i sin(2π/3)) = 4(−1/2 + (√3/2)i) = −2 + 2√3 i.

What is Arg(−5)? What is Arg(3i)?

Arg(−5) = π, because the principal range includes π but not −π. Arg(3i) = π/2.

Why can't you just use tan⁻¹(b/a) for Arg(a + bi)?

tan⁻¹ only returns angles in (−π/2, π/2), so it gives the wrong quadrant whenever a < 0. Sketch the point first, then adjust the reference angle to the correct quadrant.

State the modulus identities for a product and a quotient.

|z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂| for z₂ ≠ 0. Moduli multiply and divide.

How does Arg(z₁z₂) relate to Arg z₁ and Arg z₂?

Arg(z₁z₂) = Arg z₁ + Arg z₂ up to a multiple of 2π; add or subtract 2π to bring the sum back into (−π, π].

Prove that |z̄| = |z|.

If z = a + bi then z̄ = a − bi and |z̄| = √(a² + (−b)²) = √(a² + b²) = |z|. Geometrically, conjugation is a reflection in the real axis, which preserves distance from O.

If Arg z = θ with 0 < θ < π, what is Arg(z̄) and Arg(−z)?

Arg(z̄) = −θ (reflection in the real axis). Arg(−z) = θ − π (a half-turn about O), which lies in (−π, 0).

Simplify 6 cis(5π/6) ÷ 2 cis(−π/3).

Divide moduli and subtract arguments: 3 cis(5π/6 + π/3) = 3 cis(7π/6) = 3 cis(−5π/6) as a principal argument.

State de Moivre's theorem for integral powers.

For any integer n, (r cis θ)ⁿ = rⁿ cis(nθ). It is proved by induction for positive n and extended to negative n using 1/cis θ = cis(−θ).

Evaluate (cis(π/12))⁸.

cis(8π/12) = cis(2π/3) = −1/2 + (√3/2)i.

Evaluate (1 + i)⁻⁴.

1 + i = √2 cis(π/4), so (1 + i)⁻⁴ = (√2)⁻⁴ cis(−π) = (1/4)(−1) = −1/4.

How do you use de Moivre to express cos 3θ in terms of cos θ?

Expand (cos θ + i sin θ)³ and equate real parts with cos 3θ: cos 3θ = cos³θ − 3 cos θ sin²θ = 4cos³θ − 3 cos θ.

Geometrically, how is z₁ + z₂ found on an Argand diagram?

By vector addition: complete the parallelogram on O, z₁ and z₂; the fourth vertex is z₁ + z₂. The difference z₁ − z₂ is the vector from z₂ to z₁.