Practice Maths

← Further Complex Numbers › Exponential Form and Euler’s Formula › Solutions

Exponential Form and Euler’s Formula — Full Worked Solutions

  1. Convert between forms. Fluency

    z = 3eiπ/4

    By Euler’s formula: eiπ/4 = cos(π/4) + i sin(π/4).

    So z = 3(cos(π/4) + i sin(π/4)) = 3 cis(π/4) — this is the polar form.

    Real part: 3 cos(π/4) = 3 × √2/2 = 3√2/2

    Imaginary part: 3 sin(π/4) = 3 × √2/2 = 3√2/2

    So z = 3√2/2 + i(3√2/2).

  2. Convert to exponential form. Fluency

    z = 2(cos(π/6) + i sin(π/6))

    By Euler’s formula, cosθ + i sinθ = eiθ, so with θ = π/6:

    z = 2eiπ/6

    The modulus is r = 2 and the argument is θ = π/6.

  3. Multiply in exponential form. Fluency

    z1 = 2eiπ/3, z2 = 3eiπ/6

    z1z2 = (2 × 3) ei(π/3 + π/6) = 6ei(2π/6 + π/6) = 6eiπ/2

    Converting to a + bi: 6eiπ/2 = 6(cos(π/2) + i sin(π/2)) = 6(0 + i × 1) = 6i

  4. Divide in exponential form. Fluency

    z1 = 4ei2π/3, z2 = 2eiπ/6

    z1/z2 = (4/2) ei(2π/3 − π/6) = 2 ei(4π/6 − π/6) = 2 ei3π/6 = 2eiπ/2

    Verification: modulus = 4/2 = 2 ✓, argument = 2π/3 − π/6 = π/2 ✓

  5. Euler’s identity. Understanding

    Proof of eiπ + 1 = 0:

    By Euler’s formula: eiθ = cosθ + i sinθ.

    Set θ = π: eiπ = cos(π) + i sin(π) = −1 + i × 0 = −1.

    Therefore eiπ + 1 = −1 + 1 = 0. ✓

    Why it is beautiful: The equation eiπ + 1 = 0 unites the five most fundamental constants in all of mathematics — e (natural growth), i (the imaginary unit), π (geometry of circles), 1 (the multiplicative identity), and 0 (the additive identity) — in a single, elegant equation using only the basic operations of addition, multiplication, and exponentiation.

  6. Multiple angle formulas. Understanding

    We use ei3θ = (eiθ)3.

    LHS: ei3θ = cos(3θ) + i sin(3θ).

    RHS: (cosθ + i sinθ)3. Expand using the binomial theorem:

    = cos3θ + 3cos2θ(i sinθ) + 3cosθ(i sinθ)2 + (i sinθ)3

    = cos3θ + 3i cos2θsinθ − 3cosθsin2θ − i sin3θ

    Equating real parts: cos(3θ) = cos3θ − 3cosθsin2θ = 4cos3θ − 3cosθ

    (using sin2θ = 1 − cos2θ)

    Equating imaginary parts: sin(3θ) = 3cos2θsinθ − sin3θ = 3sinθ − 4sin3θ

    (using cos2θ = 1 − sin2θ)

  7. Modulus and argument of a complex exponential. Understanding

    z = e2+iπ/4 = e2 ⋅ eiπ/4

    Since e2 is a positive real number and eiπ/4 = cos(π/4) + i sin(π/4):

    This is in the form reiθ with r = e2 and θ = π/4.

    |z| = e2 (approximately 7.389)

    arg(z) = π/4

    In Cartesian form: z = e2(cos(π/4) + i sin(π/4)) = e2(√2/2 + i√2/2) = e2√2/2 + ie2√2/2.

  8. Verify De Moivre’s theorem. Understanding

    Step 1 — Show (eiθ)n = einθ:

    By the index law (am)n = amn, which holds for complex exponentials:

    (eiθ)n = eiθ ⋅ n = einθ ✓

    Step 2 — Verify De Moivre for z = r cisθ:

    Write z = reiθ (this is the same as r cisθ).

    zn = (reiθ)n = rn(eiθ)n = rneinθ

    Converting back: rneinθ = rn(cos(nθ) + i sin(nθ)) = rn cis(nθ).

    Therefore zn = rn cis(nθ), which is De Moivre’s theorem. ✓

  9. Power reduction using Euler’s formula. Problem Solving

    We use cosθ = (eiθ + e−iθ)/2.

    cos2θ = [(eiθ + e−iθ)/2]2

    = (eiθ + e−iθ)2 / 4

    Expand the numerator:

    (eiθ + e−iθ)2 = e2iθ + 2eiθe−iθ + e−2iθ

    = e2iθ + 2e0 + e−2iθ   [since eiθe−iθ = e0 = 1]

    = e2iθ + e−2iθ + 2

    = 2cos(2θ) + 2   [using cos(2θ) = (e2iθ+e−2iθ)/2]

    Therefore: cos2θ = [2cos(2θ) + 2] / 4 = ½(1 + cos(2θ))

    This is the standard double-angle power-reduction identity.

  10. Unit circle and Euler’s formula. Problem Solving

    Since |z| = 1, write z = eiθ = cosθ + i sinθ.

    The conjugate is z̅ = e−iθ = cosθ − i sinθ.

    z + z̅ = (cosθ + i sinθ) + (cosθ − i sinθ) = 2cosθ.

    Setting z + z̅ = √3:

    2cosθ = √3

    cosθ = √3/2

    For θ ∈ [0, 2π): θ = π/6   or   θ = 2π − π/6 = 11π/6.

    Solutions: θ = π/6 and θ = 11π/6

    Corresponding complex numbers: z = eiπ/6 = √3/2 + i/2   and   z = ei11π/6 = √3/2 − i/2.