Find all solutions for $z^3 = -27$

Question

I would like some feedback on my thought process to make sure it is logically sound. The way I think I am deriving the solutions is trigonometrically(rather then getting the roots).

The first step is to write $z^3 = -27$ in polar form, $|z|(cos\theta + i\sin\theta)$. We have to put this into polar form because it is the essential bridge between trigonometric functions and the complex field.

$$|z^3|(\cos 3\theta + i\sin 3\theta) = 27(1 + 0i) ~~~~~~~~~~~~~~~~~~~~[1]$$

My understanding starts to break down here. I observe (but don't fully understand) that

$$ \begin{align} 3\theta &= \pi + n2\pi && [2]\\ \theta &= \frac{\pi}{3} + \frac{n2\pi}{3} \end{align}$$

So the solutions are $ \theta = \frac{\pi}{3} + \pi + \frac{5\pi}{3}$. First, am I correct in my observations and understanding up to this point? If so....

I know $\theta$ is the angle measured counterclockwise from the x-axis to the vector from $0\to Z$ and $\theta = \pi + n(2\pi)$. Where does $\pi$ originate from in the $\theta$ equation? Or is that not even correct to begin with? Probably not by coincidence, I also observe that $\cos\pi = -1$ and $\sin\pi = 0$, which looks eerily related to the polar form defined above in [1].

Answer 1

To say where is $\pi$ originated from and to solve the coincidence in a reasonable manner, we first need to solve the equation on $\Bbb C$ precisely. The equation states that $$z^3=-27$$ If we assume to solve the equation over $\Bbb C$, then any complex number $z$ has a magnitude (the radius of the geometric representation with respect to the origin) and an angle (which is measured counterclockwise). In a polar representation, we have $$z=re^{i\theta}$$plugging this in the equation we obtain$$r^3e^{3i\theta}=27e^{i\pi}$$which leads to $$r^3=27\\e^{3i\theta}=e^{i\pi}$$from which we can say $$r=3\\3i\theta=\pi+2k\pi$$The last equation is true because $\sin$ and $\cos$ are periodic with a period of $2\pi$ and so is $e^{i\theta}=\cos \theta+i\sin\theta$. Finally $$\theta={2k\pi +\pi \over 3}$$this is where the $\pi$ comes from.

Answer 2

If you write $z$ as $\rho\bigl(\cos(\theta)+\sin(\theta)i\bigr)$, then$$z^3=\rho^3\bigl(\cos(\theta)+\sin(\theta)i\bigr)^3=\rho^3\bigl(\cos(3\theta)+\sin(3\theta)i\bigr).$$Since $-27=27\bigl(\cos(\pi)+\sin(\pi)i\bigr)$, you should solve the equation$$\rho^3\bigl(\cos(3\theta)+\sin(3\theta)i\bigr)=27\bigl(\cos(\pi)+\sin(\pi)i\bigr).$$So, you take $\rho=3$, and you choose $\theta$ such that $3\theta$ is of the form $\pi+2n\pi$, for some integer $n$. The possible values of $\theta$ are the ones that you got: $\frac\pi3$, $\pi$, and $\frac{5\pi}3$.

Answer 3

Another way to solve it: $z^3+27=(z+3)(z^2-3z+9)=0$ $\iff z+3=0 $ or $z^2-3z+9=0$ The solutions will be $z=-3, \frac{3+3i\sqrt{3}}{2}, \frac{3-3i\sqrt{3}}{2}$

Answer 4

Use exponential form, it's simpler to use. We know that $\;-1=\mathrm e^{i\pi}$, so $\;27=27\mathrm e^{i\pi}$ That's where the $\pi$ comes from.

This being said, $z^3=-27$ has an obvious solution: $\;z=-3$, and we know all $n$-th roots of a complex number are obtained from any of them, multiplying it by the $n$-th roots of unity. In the present case, $n=3$, and the 3rd roots of unity are the numbers $$1, \quad j=\mathrm e^{2i\pi/3}\quad\text{and}\quad j^2=\bar j=\mathrm e^{-2i\pi/3},$$ so the solutions to the given equation are $$-3,\quad-3j=3\mathrm e^{5i\pi/3},\quad -3j^2=3\mathrm e^{i\pi/3}.$$