Complex Numbers in Polar Form and Cube Roots

Complex numbers in polar form are central to HSC Extension 2, enabling elegant calculations with powers and roots. This problem takes you through the key skills: converting between Cartesian and polar forms, applying de Moivre’s theorem, and finding cube roots. By the end, you’ll see how these ideas fit together to simplify seemingly messy expressions.

Problem Statement

Suppose that $z = 4\sqrt{3}\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) - 4\left(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\right)$.

a) Simplify $z$, writing your answer in polar form. b) Show that $\frac{z}{8} + i\left(\frac{z}{8}\right)^2 + \left(\frac{z}{8}\right)^3 = 2i$. c) Find the three cube roots of $z$ in polar form.

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Hints

• For part (a), convert the polar forms to Cartesian form to perform the subtraction, then convert the result back to polar form.
• For part (b), use the polar form from part (a) to compute powers of $\frac{z}{8}$ using de Moivre's theorem.
• For part (c), apply the formula for the $n$-th roots of a complex number: $w_k = r^{1/n}\text{cis}\left(\frac{\theta + 2k\pi}{n}\right)$.

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Solutions

• Part (a): Because addition and subtraction of complex numbers are simplest in Cartesian form, we first convert each term from polar to Cartesian using the exact values of cosine and sine.

  First, expand both terms into Cartesian form:

  $$4\sqrt{3}\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} + 6i$$

  $$4\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -2\sqrt{3} + 2i$$

  Now, subtract the second term from the first:

  $$z = (2\sqrt{3} + 6i) - (-2\sqrt{3} + 2i) = 4\sqrt{3} + 4i$$

  Now that have $z$ in Cartesian form, we find its modulus and argument to express it in polar form. The modulus gives the distance from the origin, and the argument is the angle from the positive real axis.

  To write this in polar form, find the modulus $r$ and argument $\theta$:

  $$r = |z| = \sqrt{(4\sqrt{3})^2 + 4^2} = \sqrt{48 + 16} = \sqrt{64} = 8$$

  $$\theta = \arg(z) = \tan^{-1}\left(\frac{4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$

  Therefore, $z = 8\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)$.

• Part (b): From part (a), we know $\frac{z}{8} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$. Since this is a unit-modulus complex number, de Moivre's theorem tells us that raising it to a power $n$ simply multiplies its argument by $n$.

  Using de Moivre's theorem, we can find the powers:

  $$\left(\frac{z}{8}\right)^2 = \cos \frac{2\pi}{6} + i \sin \frac{2\pi}{6} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$

  $$\left(\frac{z}{8}\right)^3 = \cos \frac{3\pi}{6} + i \sin \frac{3\pi}{6} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = i$$

  Now substitute these into the given expression:

  $$\frac{z}{8} + i\left(\frac{z}{8}\right)^2 + \left(\frac{z}{8}\right)^3 = \left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) + i\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) + i$$

  Distributing the $i$ and grouping real and imaginary parts makes the simplification clear:

  $$= \frac{\sqrt{3}}{2} + \frac{1}{2}i + \frac{1}{2}i - \frac{\sqrt{3}}{2} + i$$

  $$= \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} + \frac{1}{2} + 1\right)i = 2i$$

• Part (c): We now find the three distinct complex numbers whose cube is $z$. The general $n$th root formula for $r\text{cis}\theta$ is $r^{1/n}\text{cis}\frac{\theta+2k\pi}{n}$ for $k=0,1,\ldots,n-1$. Here $n=3$, $r=8$, and $\theta = \frac{\pi}{6}$.

  We need to find the cube roots of $z = 8\text{cis}\frac{\pi}{6}$. Using the root formula, the roots $w_k$ are given by:

  $$w_k = 8^{1/3}\text{cis}\left(\frac{\frac{\pi}{6} + 2k\pi}{3}\right) \text{ for } k = 0, 1, 2$$

  For $k=0$:

  $$w_0 = 2\text{cis}\left(\frac{\pi}{18}\right)$$

  For $k=1$:

  $$w_1 = 2\text{cis}\left(\frac{\frac{\pi}{6} + 2\pi}{3}\right) = 2\text{cis}\left(\frac{13\pi}{18}\right)$$

  For $k=2$:

  $$w_2 = 2\text{cis}\left(\frac{\frac{\pi}{6} + 4\pi}{3}\right) = 2\text{cis}\left(\frac{25\pi}{18}\right) = 2\text{cis}\left(-\frac{11\pi}{18}\right)$$

  The last root has been expressed with a principal argument in the interval $(-\pi,\pi]$.

  So the three roots are $2\left(\cos\frac{\pi}{18} + i\sin\frac{\pi}{18}\right)$, $2\left(\cos\frac{13\pi}{18} + i\sin\frac{13\pi}{18}\right)$, and $2\left(\cos\left(-\frac{11\pi}{18}\right) + i\sin\left(-\frac{11\pi}{18}\right)\right)$.

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Takeaways

• Converting between polar and Cartesian forms is a fundamental skill for simplifying complex expressions involving both addition/subtraction and powers/roots.
• De Moivre's theorem provides a highly efficient way to compute powers of complex numbers in polar form.
• The $n$-th roots of a complex number are evenly spaced around a circle in the complex plane, separated by an angle of $\frac{2\pi}{n}$.

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Further Readings

HSC Complex Numbers, HSC Last Resorts