Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
A point in the complex plane can be described by its distance and angle as well as by its coordinates, just like polar coordinates. For complex numbers this polar description makes multiplication remarkably simple, as the next lesson shows.
🔑 Idea clave
$$z = a + bi = r(\cos\theta + i\sin\theta),$$ where $r = |z| = \sqrt{a^2 + b^2}$ is the modulus and $\theta$ is the argument, the angle from the positive real axis (with $\tan\theta = \frac{b}{a}$ in the right quadrant). Going back: $a = r\cos\theta$, $b = r\sin\theta$. Some books write $r\operatorname{cis}\theta$ for short.
💡 Por qué funciona
$z$ is the point $(a, b)$. Its polar coordinates are $(r, \theta)$ with $a = r\cos\theta$ and $b = r\sin\theta$, so $$a + bi = r\cos\theta + ir\sin\theta = r(\cos\theta + i\sin\theta).$$
With polar marks on, the readout gives the modulus and argument of $z$.
Worked examples
✏️ Ejemplo resuelto: To polar form
Write $-1 + i\sqrt{3}$ in polar form.
$r = \sqrt{1 + 3} = 2$.
The point $(-1, \sqrt{3})$ is in Quadrant II with reference angle $\frac{\pi}{3}$ (since $\tan = \sqrt{3}$), so $\theta = \frac{2\pi}{3}$.
Reading the argument off a calculator's $\tan^{-1}\frac{b}{a}$ without checking the quadrant, the same trap as with polar coordinates. For $-1 + i\sqrt{3}$ that gives $-\frac{\pi}{3}$, which points into Quadrant IV. Plot the point first.
Práctica
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