= Solution
Identify $\mathbb C$ with the Euclidean plane and start with the points $0,1$. A complex number is a <constructible number> if a finite sequence of drawing lines through known points and circles with known centres and known distances as radii, and selecting their intersections, produces that point.
Line-line intersections solve linear equations in the current coordinate <field>. Line-circle and circle-circle intersections require at most one quadratic equation after eliminating a coordinate. Hence all coordinates obtained lie in a finite tower of real quadratic extensions of $\mathbb Q$. If $\alpha=x+iy$ is constructible, choose a common such real <field> $F$ containing $x,y$. Adjoining $i$ if needed still gives an extension $F(i)/\mathbb Q$ of power-of-two degree. By the <tower law for field extensions>,
$$
\boxed{[\mathbb Q(\alpha):\mathbb Q]\text{ is a power of }2.}
$$
A regular heptagon centred at $0$ with one vertex $1$ would have $\xi=e^{2\pi i/7}$ as another constructible vertex. But its cyclotomic degree is $\varphi(7)=6$, which is not a power of two. Thus \b[the regular heptagon is not constructible].
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