For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane iff $\Re(w)=-\frac12$. For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane if, and only if, $\Re(w)=-\frac12$.
I came across this fact while messing about with spirals of powers of a complex number. I've tried parametrising the straight line with a real number $t$, so, for example, $z(t)=(1-t)+tw$. But alas to no avail! Any insights are appreciated.
Here is a link to a Desmos graph showcasing the phenomenon:
https://www.desmos.com/calculator/gsugcdlkc9
 A: $1,\omega,\omega^3$ are collinear if and only if $$\frac{\omega^3-\omega}{\omega-1}$$ is real. Letting $\omega=x+yi $, the condition boils down to $$(2x+1)y=0,$$ hence your result.
Edit: More precisely, the OP can say that $1,\omega,\omega^3$ are collinear if and only $\omega$ is real or $x=-\frac 12$.
A: Continue with the parametrized line you have $z=(1-t)+tw$ and substitute $w^3$ into it to get 
$$w^3-1 = t(w-1)$$
Assume $w\ne1$, the equation reduces to 
$$w^2+w+1=t$$
Since $t$ is real, $t-\bar{t}=0$, which leads to $(w^2+w+1)-(\bar{w}^2+\bar w+1)=0$, or 
$$(w-\bar{w})(w+\bar w +1)=0$$
With $w-\bar w\ne 0$,
$$w+\bar w +1 = 2Re(w)+1=0\implies Re(w) = -\frac12$$
The reverse is true as well. 
A: You want $\omega^3=t(\omega-1)+1\implies \omega^3-t\omega-(1-t)=0$.  So you need to solve a cubic.  
Sure enough, $\omega=1$ is a solution.  Factor: $(\omega-1)(\omega^2+\omega+(1-t))$.  
So solve the quadratic factor: $\omega=\dfrac{-1\pm\sqrt{1-4(1)(1-t)}}2\implies\omega=\dfrac{-1\pm\sqrt{-3+4t}}2$.
At any rate, your result appears to be incorrect.
We get solutions for $\omega$ varying with $t$.
For $t\le3/4$, we do get the line $\Re(\omega)=-1/2$ included among the solutions.
Otherwise $\omega $ can be real.
So you can correct your statement by saying that if $\Im(\omega)\ne0$, then $\Re(\omega)=-1/2$.
A: A different proof:
Let $\omega = x + iy\in\mathbb{C}, y \neq 0, \omega \neq 1$
$\omega^3 = (x^3 - 3xy^2) + i(3x^2y - y^3)$
The three points $\omega \equiv(x, y)$, $(1,0)$ and $\omega^3 \equiv (x^3 - 3xy^2, 3x^2y - y^3)$ are collinear, iff the following determinant is $0$
\begin{vmatrix}
x & y & 1 \\ 
1 & 0 & 1 \\ 
x^3 - 3xy^2 & 3x^2y - y^3 & 1 
\end{vmatrix} 
Simplifying,
$y(2x+1)[(x-1)^2 + y^2)] = 0$
Since $y \neq 0$, this is possible iff $ x = -\frac{1}{2}$
A shorter proof:
Let $\omega=x + iy, y \neq 0, \omega \neq 1$
$1,\omega,\omega^3$ are collinear iff $\frac{\omega^3-\omega}{\omega-1} = \omega^2 + \omega + 1$ is real.
Since $\omega^2 + \omega + 1 = (x^2 + x - y^2 + 1) + i y(2x + 1)$, we must have $y(2x + 1) = 0$
Now $y \neq 0 \implies x = -\frac{1}{2}$ 
