A question from my text asks to find the $3$ cube roots of $-27$. The first step in the solution is to immediately show the polar form of $-27$ as
$$-27 = 27(\cos \pi + i\sin \pi).$$
Would someone mind explaining to me the jump from $-27$ to its polar form as shown by the text?
On
That's the simplest way of solving your problem.
In general, if you have an equation
$$z^n = w$$
then the simplest way to solve this equation is to convert $w$ into polar coordinates, i.e.
$$w=r (\cos\phi + i\sin \phi)$$
then every solution of $z^n=w$ is $$\sqrt[n]{r} \left((\cos\left(\frac{\phi}{n} + \frac{2k\pi}{n}\right)+i\sin\left(\frac{\phi}{n} + \frac{2k\pi}{n}\right)\right)$$
for $k=0,1,2\dots,n-1$
Sure. Any complex number in Cartesian form $a+ib$ can be represented in polar form as $r \text{ cis } \theta$, where $\text{cis } \theta$ is shorthand for $\cos \theta + i \sin \theta$ (hence the acronym). In this form, $r$ represents the magnitude of the number—its distance from the origin—and $\theta$ represents its argument. The argument of a complex number is the angle that a line from the origin to it makes with the positive $x$-axis (moving counter-clockwise).
The distance of $-27$ from the origin is simply $27$, so $r = 27$. A line from the origin to $-27$ goes in the negative $x$ direction, so the angle made with the positive $x$-axis is $\pi$ radians, so $\theta = \pi$. Thus, $-27 = 27 \text{ cis } \pi$.
More generally, the distance of $a+ib$ from the origin is $r = \sqrt{a^2+b^2}$, and the argument of $a+ib$ is $\theta = \tan^{-1} b/a$ if $b > 0$, and $\theta = -\tan^{-1} b/a$ if $b < 0$. If $b = 0$, then $\theta = 0$ if $a > 0$, and $\theta = \pi$ if $a < 0$. If $a = b = 0$, then the polar representation is just $r = 0$.