Complex Numbers and De Moivre's Theorem

Tutorial 2: Complex Numbers and De Moivre's Theorem

1. Cartesian Form for Powers

Given $z_1 = -4 + 4i$ and $z_2 = -2\sqrt{3} + 2i$

  • $(z_1)^3$:

    • Polar form: $r_1 = \sqrt{(-4)^2 + 4^2} = 4\sqrt{2}$, $\theta_1 = \frac{3\pi}{4}$
    • $z_1 = 4\sqrt{2} \left( \cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4} \right)$
    • $(z_1)^3 = (4\sqrt{2})^3 \left( \cos\frac{9\pi}{4} + i \sin\frac{9\pi}{4} \right) = 128\sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right)$
    • Cartesian: $128\sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 128 + 128i$

  • $(z_2)^4$:

    • Polar form: $r_2 = \sqrt{(-2\sqrt{3})^2 + 2^2} = 4$, $\theta_2 = \frac{5\pi}{6}$
    • $z_2 = 4 \left( \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6} \right)$
    • $(z_2)^4 = 4^4 \left( \cos\frac{20\pi}{6} + i \sin\frac{20\pi}{6} \right) = 256 \left( \cos\frac{10\pi}{3} + i \sin\frac{10\pi}{3} \right)$
    • Cartesian: $256 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -128 + 128\sqrt{3}i$

2. Simplify to Terms in $\cos x$ and $\sin x$

Use: $e^{i\theta} = \cos\theta + i\sin\theta$

(a) $(\cos 7x + i \sin 7x)(\cos 5x - i \sin 5x)$

  • $= e^{i7x} \cdot e^{-i5x} = e^{i2x} = \cos 2x + i \sin 2x$

(b) $\frac{(\cos 3x + i \sin 3x)^3}{(\cos 4x + i \sin 4x)^2(\cos 2x - i \sin 2x)}$

  • Numerator: $(e^{i3x})^3 = e^{i9x}$
  • Denominator: $(e^{i4x})^2 \cdot e^{-i2x} = e^{i8x} \cdot e^{-i2x} = e^{i6x}$
  • Result: $e^{i9x} / e^{i6x} = e^{i3x} = \cos 3x + i \sin 3x$

(c) $\frac{(\cos 2x - i \sin 2x)^4}{(\cos 5x + i \sin 5x)^3}$

  • Numerator: $(e^{-i2x})^4 = e^{-i8x}$
  • Denominator: $(e^{i5x})^3 = e^{i15x}$
  • Result: $e^{-i8x} / e^{i15x} = e^{-i23x} = \cos 23x - i \sin 23x$

3. Exponential and Cartesian Forms

(a) $2\left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right)$

  • Exponential: $2e^{i\pi/4}$
  • Cartesian: $2\left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = \sqrt{2} + i\sqrt{2}$

(b) $\left[ \frac{1}{2} \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right) \right]^6$

  • $= \left( \frac{1}{2} \right)^6 e^{i6\pi/3} = \frac{1}{64} e^{i2\pi} = \frac{1}{64} \cdot 1$
  • Exponential: $\frac{1}{64} e^{i0}$
  • Cartesian: $\frac{1}{64}$

4. Modulus and Argument Calculations

(a) $\arg \left[ \left( \cos\frac{\pi}{4} - i \sin\frac{\pi}{4} \right)^3 \left( \cos\frac{2\pi}{3} + i \sin\frac{2\pi}{3} \right)^4 \right]$

  • First factor: $(e^{-i\pi/4})^3 = e^{-i3\pi/4}$
  • Second factor: $(e^{i2\pi/3})^4 = e^{i8\pi/3} = e^{i2\pi/3}$
  • Product: $e^{-i3\pi/4} \cdot e^{i2\pi/3} = e^{-i\pi/12}$
  • Argument: $-\frac{\pi}{12}$

(b) Modulus and argument of $\frac{(1 + i \sqrt{3})^4}{(1 + i)^6}$

  • Numerator: $z_1 = 1 + i\sqrt{3} = 2e^{i\pi/3}$, so $z_1^4 = 16e^{i4\pi/3}$
  • Denominator: $z_2 = 1 + i = \sqrt{2} e^{i\pi/4}$, so $z_2^6 = (\sqrt{2})^6 e^{i6\pi/4} = 8e^{i3\pi/2}$
  • Result: $\frac{16}{8} e^{i(4\pi/3 - 3\pi/2)} = 2e^{-i\pi/6}$
  • Modulus: 2
  • Argument: $-\frac{\pi}{6}$

5. Euler's Formula and Trigonometric Identities

Given: $e^{ix} + e^{-ix} = 2\cos x$ and $e^{ix} - e^{-ix} = 2i \sin x$

  • (a) $e^{inx} + e^{-inx} = 2\cos(nx)$
  • (b) $e^{inx} - e^{-inx} = 2i \sin(nx)$

For $\cos 4x$ and $\sin 4x$:

  • $e^{i4x} = (\cos x + i \sin x)^4$
  • Expand: $\cos 4x + i \sin 4x = \cos^4 x - 6\cos^2 x \sin^2 x + \sin^4 x + i(4\cos^3 x \sin x - 4\cos x \sin^3 x)$
  • $\cos 4x = \cos^4 x - 6\cos^2 x \sin^2 x + \sin^4 x$
  • Using $\sin^2 x = 1 - \cos^2 x$: $\cos 4x = 8\cos^4 x - 8\cos^2 x + 1$
  • $\sin 4x = 4\cos^3 x \sin x - 4\cos x \sin^3 x$

6. Prove Trigonometric Identities

Using De Moivre's Theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

(a) $\cos(5\theta) = 16\cos^5 \theta - 20\cos^3 \theta + 5\cos\theta$

  • $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i \sin 5\theta$
  • Expand LHS using binomial theorem:
    • Real part: $\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$
  • Substitute $\sin^2\theta = 1 - \cos^2\theta$ and simplify to get $16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$

(b) $\sin(5\theta) = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$

  • Imaginary part of expansion: $5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$
  • Substitute $\cos^2\theta = 1 - \sin^2\theta$ and simplify to get $16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$

7. Cube Roots

(

Source: FAC1004 Tutorial 2 25-26.pdf