Revise Complex numbers: De Moivre's theorem and roots for IB Mathematics: Analysis and Approaches (1.7)

In polar form a complex number carries a modulus $r$ (its distance from the origin) and an argument $\theta$ (its angle). Multiplying two complex numbers multiplies their moduli and adds their arguments. Multiplying $z$ by itself $n$ times therefore raises the modulus to the power $n$ and multiplies the argument by $n$. That single observation is De Moivre's theorem.

If $z = r(\cos\theta + i\sin\theta)$ and $n$ is an integer, then $z^n = \big[r(\cos\theta + i\sin\theta)\big]^n = r^n(\cos n\theta + i\sin n\theta).$ Using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$, the same statement is $\big(re^{i\theta}\big)^n = r^n e^{in\theta}.$

To solve $z^n = w$ we run De Moivre in reverse. A real number such as $4$ has two square roots ($\pm 2$); more generally every non-zero complex number has exactly $n$ distinct $n$th roots. The trick is to remember that an angle is only defined up to whole turns: $w = re^{i\theta}$ is the same number as $re^{i(\theta + 2\pi k)}$ for any integer $k$. When we take the $n$th root and divide the argument by $n$, those extra turns become $n$ genuinely different angles.

The $n$ distinct $n$th roots of $w = re^{i\theta}$ are $z_k = r^{1/n}\, e^{\,i\left(\frac{\theta + 2\pi k}{n}\right)}, \qquad k = 0, 1, 2, \dots, n-1.$ Equivalently, $z_k = r^{1/n}\left(\cos\dfrac{\theta + 2\pi k}{n} + i\sin\dfrac{\theta + 2\pi k}{n}\right)$.

A special and important case is $z^n = 1$, whose solutions are the $n$th roots of unity. Since $1 = e^{i\cdot 0}$ with modulus $1$, the roots are $z_k = e^{i\,2\pi k/n}$ for $k = 0, 1, \dots, n-1$. They all sit on the unit circle, equally spaced $\frac{2\pi}{n}$ apart, and one of them is always $z = 1$ (the case $k=0$). For $n \ge 2$ the roots of unity sum to zero, because they are symmetrically placed around the origin. Writing $\omega = e^{i\,2\pi/n}$, every root is a power of this one: $1, \omega, \omega^2, \dots, \omega^{n-1}$.

De Moivre's theorem is also a bridge to trigonometry. Expanding $(\cos\theta + i\sin\theta)^n$ with the binomial theorem and comparing it with $\cos n\theta + i\sin n\theta$ produces formulae for $\cos n\theta$ and $\sin n\theta$ in powers of $\cos\theta$ and $\sin\theta$.

The $n$th roots of any complex number share one modulus, $r^{1/n}$, so they all lie on a circle of that radius centred at the origin. Their arguments increase in equal steps of $\frac{2\pi}{n}$, so they are evenly spaced and form the vertices of a regular $n$-sided polygon. This symmetry is a free checking tool: cube roots make an equilateral triangle, fourth roots make a square, fifth roots make a regular pentagon. If your plotted roots are not evenly spaced, you have made an arithmetic slip — almost always in an argument.

IB Mathematics marks Paper 1 analytically. Each mark is tied to a specific line of working: an M mark rewards a correct method, and an A mark rewards accuracy but is dependent on the method — you cannot earn the A without the M it hangs from. The engine accepts any exact or equivalent form, so $2e^{i\,2\pi/3}$ and $2e^{i\,4\pi/3}$ (or their negatives-argument or Cartesian equivalents) all score. Study how each mark below is earned by a single line.

De Moivre's theorem is the engine behind much of the complex-number toolkit. Solving $z^n = w$ underlies polynomial factorisation over the complex numbers, since the roots of unity generate the factors of $z^n - 1$. The same power-and-turn picture reappears when complex numbers model oscillations and rotations, and the multiple-angle identities you derived here feed directly into trigonometry and calculus. Master the habit — convert to polar or Euler form, stretch the modulus, multiply or divide the argument, and for roots share the turn into $n$ equal slices — and these problems become variations on a method you already own.