Find the complex roots of $8z^3+27=0$ and thus find the area of the triangle

Question

$8z^3+27=0$

So I took two approaches, and my answers are slightly different, so I was wondering which one was correct: \begin{align*} 8z^3 & =-27\\ 2z & = -3\\ z & = -\dfrac32 \end{align*}

$r=\dfrac32$

$\arg z = \arctan\left(\dfrac{0}{-3/2}\right) = 0$

$z = \dfrac32\operatorname{cis}\!\big(0 + 2\pi k\big)\,$, $\;\,k$ is a positive integer

second approach

$z^3 = -\dfrac{27}8$

$r=\dfrac{27}8$

$\arg z = 0$

$z^{1/3} = \left(\dfrac{27}8\right)^{\!1/3}\!\operatorname{cis}\left(\dfrac{0 +2\pi k}{3}\right)$

$z =\dfrac32\operatorname{cis}\left(\dfrac{2 \pi k}{3}\right)$

Area of the triangle:
$\dfrac12ab\sin(c)=\dfrac12\left(\dfrac32\right)^{\!2}\!\sin\left(\dfrac{2\pi}3\right)\approx0.041$

Answer 1

Let $z=Re^{i\theta}$, where $R\in[0,\infty)$ and $\theta\in[0,2\pi)$, Then, $$R^3e^{i\cdot3\theta} = -\dfrac{27}{8}\Rightarrow R=\dfrac{3}{2} \enspace \text{and} \enspace \theta \in \left\{\dfrac{\pi}{3},\pi,\dfrac{5\pi}{3}\right\}$$ Hence, the solution is $\dfrac{3}{2}e^{i\frac{\pi}{3}}, \dfrac{3}{2}e^{i\pi}\enspace \text{or} \enspace \dfrac{3}{2}e^{i\frac{5\pi}{3}}$.

Answer 2

A more usual notation for cis(x) is $e^{ix}.$

The second approach is ok but you can restrict explicitely the values of $k$ to three consecutive integers (e.g. $-1,0,1$), to avoid redundancy.

The first one is not, mainly because a non-zero complex number has three cube roots, not only one, so you missed 2 solutions out of 3. (Also, as noted by @N.F.Taussig, "$\arg z = \arctan(\frac{0}{-3/2}) = \pi$" makes no sense. Replace it by "$\arg z = \arg(-3/2) = \pi$". And "$k$ is a positive integer" can be replaced by $k=0$ without missing more solutions.)

Note that the $z=\frac32e^{i(1+2k)\pi}$ of your first approach are the particular case of the $z=\frac32e^{i\frac{1+2\ell}3\pi}$ of your second approach when $ℓ=3k+1.$ Since (as you can check by cubing them, following @CyclotomicField's advice) the latter are true solutions, this is another way to realize that former are not all the solutions.