Roots of complex number

Question

Solve the equation $$z^3-1=0$$ Show that the roots are represented in an Argand diagram by the vertices of an equilateral triangle.

(EDITED: Thank you for your quick respond)

Answer 1

Some simple observations:

a) Since $z^3 = 1$, we know $|z^3| = |z|^3 = |1| = 1$.

This implies $|z| = 1$ (since $|z| \in \Bbb R^+_0$). So all of our roots (should they exist) lie on the unit circle in the Argand plane.

b) If $z \neq 1$, then $z$ is a root of $\dfrac{x^3 -1}{x -1} = x^2 + x + 1$. This can be solved via the quadratic equation, if one wishes to.

c) Complex multiplication is closed in the unit circle, and corresponds to rotation. Thus we seek points on the unit circle with angle $\theta$ (relative to the positive real axis), such that $3\theta =$ an integer multiple of $2\pi$. There are two such angles in the range: $\theta \in (0,2\pi)$.

Answer 2

$z^{3}=1$, using De Moivre formula we get that $z^{\frac{1}{3}}=|z|^{\frac{1}{3}} \cdot [\cos(\frac{\phi + 2 \pi r}{n})+i\sin(\frac{\phi + 2 \pi r}{n})]$, $r \in {0, 1, 2, \ldots n-1}$ $z^{\frac{1}{3}}=cos(\frac{2 \pi r}{3})+i\sin(\frac{2 \pi r}{3})$, where $r \in {0, 1, 2}$.

After this, we can draw points, corresponding to the roots, using a unit circle.

Answer 3

As a different approach, $z=1$ is a root of $z^3-1$ if and only if $(z-1)$ is a factor of $z^3-1$. Using algebra, you should verify that \begin{equation} z^3-1=(z-1)(z^2+z+1). \end{equation} Hence, $z^3-1=0$ yields $z-1=0$ (this gives the solution you were already aware of) or $z^2+z+1=0$. The second equation is quadratic and easy to find solutions for. From here, simply plot the roots you found and verify that "connecting the dots" yields an equilateral triangle.

Answer 4

$$(\rho \exp(i\theta))^3=\rho^3\exp(3i\theta)=\exp(0)$$

so $\rho=1$ and $3\theta\equiv 0 \mod 2\pi$

So $1,\exp\big(i\dfrac{2\pi}{3}\big)$ and $\exp\big(i\dfrac{4\pi}{3}\big)$ are solutions. And there are 3 solutions, so you have all three.