Graphing Complex Equation With Variable Coefficients! Let $\omega$ $=$ $-\frac{1}{2}$+$i\frac{\sqrt3}{2}$ (Where $i=\sqrt-1$).Consider S , the set of all complex numbers in the Argand Plane of the form $$a+b\omega +c \omega^2  . \forall (a,b,c) \in [0,1]$$  Plot the area covered by S

I honestly have the knowledge level of a school student in complex no. I honestly dont know how varying a,b,c will give me any shape however the solution mentions theres an hexagon, i can prove that I tried  multiplying the given equation ($a+b\omega +c \omega^2$) with its conjugate to at least get some relation since its a known result but a,b,c varying over [0,1] still makes it confusing. Will appreciate any help with this problem, thanks!

 A: Preliminaries on convex hulls of points
Here is a very useful geometric fact which doesn't really have anything to do with complex numbers, just the geometry of the plane, but I'll phrase it using complex numbers.
Fact 1: Given points $z_1,\dots,z_n\in\mathbb{C}$, the set $$\{a_1z_1+\cdots+a_nz_n\mid a_1,\dots, a_n\in[0,1]\;\textrm{and}\;a_1+\cdots+a_n=1\}$$ is the convex hull of the points $z_1,\dots,z_n$.

To get an idea for why this is true, you could think of $$\frac{a_1z_1+\cdots+a_nz_n}{a_1+\cdots+a_n}$$ as a weighted average of the points $z_1,\dots,z_n$, and the average of a collection of points has to lie "between them" in some sense.
We can use this fact to prove a closely related fact.
Fact 2: Given points $z_1,\dots,z_n\in\mathbb{C}$, the set $$\{a_1z_1+\cdots+a_nz_n\mid a_1,\dots, a_n\in[0,1]\;\textrm{and}\;a_1+\cdots+a_n\le1\}$$ is the convex hull of the points $z_1,\dots,z_n$ together with the point $0$.
Notice the only difference is we allow an inequality in the set, and now our convex hull includes the origin. To see this follows from the first fact, write $z_0=0$, and $a_0=1-(a_1+\cdots+a_n)\in[0,1]$, then $$\begin{aligned}\{a_1z_1+&\cdots+a_nz_n\mid a_1,\dots, a_n\in[0,1]\;\textrm{and}\;a_1+\cdots+a_n\le1\}=\\ &\{a_0z_0+a_1z_1+\cdots+a_nz_n\mid a_0,a_1,\dots, a_n\in[0,1]\;\textrm{and}\;a_0+a_1+\cdots+a_n=1\}.\end{aligned}$$
Solving the problem
Your expression in the question is not exactly the same as that given in the facts above, because the coefficients $a$, $b$ and $c$ could sum to more than 1. But we can easily fix this.
First assume that $a\le b\le c$, then we can rewrite $$a+b\omega+c\omega^2=a(1+\omega+\omega^2)+(b-a)(\omega+\omega^2)+(c-b)\omega^2.$$
Notice that this first term is zero since $1+\omega+\omega^2=0$, and that by our assumption $b-a,c-b\in[0,1]$. Thus rewriting $z_1=\omega+\omega^2$ and $z_2=\omega^2$, and $a_1=b-a$ and $a_2=c-b$, we can apply Fact 2 (because $a_1+a_2=c-a\le 1$. This implies that the set of points $$a+b\omega+c\omega^2\;\forall a,b,c\in[0,1]\;\textrm{such that}\;a\le b\le c$$ is the convex hull of the points $0$, $\omega+\omega^2$, and $\omega^2$.

Now if we repeat this with all 6 possible orderings of $a$, $b$, and $c$, we get 6 similar pictures which all combine to give the regular hexagon with vertices $1,1+\omega,\omega,\omega+\omega^2,\omega^2,\omega^2+1$, which is the set $S$ you were after.

