Trigonometry, Arguments, and De Moivre’s Theorem

Problem Statement

In Cartesian form ($x + iy$), multiplying two complex numbers requires tedious algebraic expansion. However, if we convert complex numbers into polar form—using their modulus ($r$) and argument ($\theta$), often written as $r(\cos\theta + i\sin\theta)$ or $r \text{cis}\theta$—multiplication becomes remarkably simple. You just multiply the moduli and add the arguments.

This trigonometric property leads to De Moivre’s Theorem: $(r \text{cis}\theta)^n = r^n \text{cis}(n\theta)$.

This theorem is the ultimate weapon for finding the roots of complex numbers and solving high-degree polynomial equations. When combined with the Complex Conjugate Root Theorem, we can completely factorise polynomials that were previously unsolvable.

Consider the polynomial equation: $z^3 = 8i$

(a) Find the modulus $r$ and the principal argument $\theta$ of the complex number $8i$. Express $8i$ in polar form ($r \text{cis}\theta$).

(b) Use De Moivre's Theorem to find the three complex roots of this equation. Express your answers in exact polar form, ensuring the arguments are principal ($-\pi < \theta \le \pi$).

(c) Convert the root in the first quadrant into Cartesian form ($x + iy$).

Hints

• Part (a): Plot $8i$ on the Argand diagram. It lies on the positive imaginary axis. Its distance from the origin is its modulus, and its angle from the positive real axis is its argument.
• Part (b): Let $z = r \text{cis}\theta$. Then $z^3 = r^3 \text{cis}(3\theta)$. Equate this to your polar form from (a). However, because angles repeat every $2\pi$, you must write the argument of $8i$ as $\frac{\pi}{2} + 2k\pi$ where $k$ is an integer. Solve for $r$ and $\theta$ by substituting $k = 0, 1, -1$ to find the three distinct roots.
• Part (c): Identify which of your three $\theta$ values lies between $0$ and $\frac{\pi}{2}$. Use the exact trigonometric ratios to evaluate $r(\cos\theta + i\sin\theta)$.

Solutions

Part (a): Polar Form Conversion

1. The complex number is $W = 0 + 8i$.
2. Plotting this on the Argand diagram, it lies on the vertical $y$-axis, 8 units above the origin.
3. Modulus: $r = \sqrt{0^2 + 8^2} = 8$.
4. Argument: The angle from the positive $x$-axis to the positive $y$-axis is $90^\circ$, or $\frac{\pi}{2}$ radians. So, $\theta = \frac{\pi}{2}$.
5. Polar form: $W = 8 \left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right) = 8 \text{cis}(\frac{\pi}{2})$.

Part (b): Finding Roots with De Moivre

1. Let the roots be $z = r \text{cis}\theta$.
2. Therefore, $z^3 = (r \text{cis}\theta)^3$. By De Moivre's Theorem, this is $r^3 \text{cis}(3\theta)$.
3. Equate this to $8i$ in general polar form (adding $2k\pi$ to account for full rotations): $r^3 \text{cis}(3\theta) = 8 \text{cis} \left( \frac{\pi}{2} + 2k\pi \right)$
4. Equate the moduli: $r^3 = 8 \implies r = 2$ (since $r$ is a real, positive distance).
5. Equate the arguments: $3\theta = \frac{\pi}{2} + 2k\pi$ $\theta = \frac{\pi}{6} + \frac{2k\pi}{3}$
6. To find the three unique roots, substitute consecutive integers for $k$ (e.g., $k=0, 1, -1$) until we have three angles within the principal domain $(-\pi, \pi]$:
   • For $k=0$: $\theta_0 = \frac{\pi}{6}$
   • For $k=1$: $\theta_1 = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{6} + \frac{4\pi}{6} = \frac{5\pi}{6}$
   • For $k=-1$: $\theta_{-1} = \frac{\pi}{6} - \frac{2\pi}{3} = \frac{\pi}{6} - \frac{4\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$
7. The three roots in polar form are:
   • $z_0 = 2 \text{cis} \left( \frac{\pi}{6} \right)$
   • $z_1 = 2 \text{cis} \left( \frac{5\pi}{6} \right)$
   • $z_2 = 2 \text{cis} \left( -\frac{\pi}{2} \right)$

Part (c): Cartesian Conversion

1. The root in the first quadrant has an angle between $0$ and $\frac{\pi}{2}$. This is $z_0 = 2 \text{cis} \left( \frac{\pi}{6} \right)$.
2. Expand the $\text{cis}$ notation: $z_0 = 2 \left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right)$
3. Evaluate the exact trigonometric ratios ($\frac{\pi}{6} = 30^\circ$):
   • $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$
   • $\sin\frac{\pi}{6} = \frac{1}{2}$
4. Substitute these values back: $z_0 = 2 \left( \frac{\sqrt{3}}{2} + i\frac{1}{2} \right)$
5. Distribute the 2: $z_0 = \sqrt{3} + i$

Takeaways

• The Roots of Unity: Solving equations like $z^n = W$ always produces $n$ distinct roots spaced equally around a circle in the complex plane. In this case, the 3 roots form an equilateral triangle on a circle of radius 2.
• The $2k\pi$ Trick: When finding complex roots, you must add $2k\pi$ to the argument before dividing by $n$. If you divide first, you will only find one root.
• Trigonometry is Essential: Finding arguments ($\tan^{-1}(y/x)$) and converting back to Cartesian coordinates requires absolute fluency in exact trigonometric ratios and quadrant signs.

Further Readings

• HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/
• HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/

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De Moivre's Theorem is a cornerstone of advanced mathematics. To master it, practice is key! Head to Vu's Maths Hub and download the HSC booklets for exhaustive practice problems. Watch how I tackle the hardest roots of unity questions on my YouTube channel. Follow my Instagram for daily maths tips, and subscribe to my Substack to join the conversation on advanced mathematical concepts.