A regular n -gon is inscribed in the circle of radius R…
A regular $n$-gon is inscribed in the circle of radius $R$ centered at $c=a+bi$, with one vertex at $z_0=a+(b+R)i$. Determine the other vertices.
Explanation of Method
Translate the center to the origin, where $z_0$ becomes the purely imaginary point $Ri$; the remaining vertices are obtained by rotating this by successive multiples of $\frac{2\pi}n$. Translate back to recover the vertices relative to $c$.
Step 1 · Verify $z_0$ lies on the circle
$$ |z_0-c|=|Ri|=R \quad\checkmark $$
Step 2 · Translate to the origin
$$ w_0=z_0-c=Ri=Re^{i\pi/2} $$
Step 3 · Rotate by multiples of $\frac{2\pi}n$
The vertices of a regular $n$-gon centered at the origin are obtained from any one vertex by rotating by $\frac{2\pi}n$ repeatedly:
$$ w_k=w_0\,e^{2\pi ik/n}=Re^{i\left(\frac\pi2+\frac{2\pi k}n\right)}, \qquad k=0,1,\ldots,n-1 $$
Step 4 · Translate back
$$ \boxed{z_k=c+Re^{i\left(\frac\pi2+\frac{2\pi k}n\right)}} $$
Step 5 · Expand into real and imaginary parts
Using $\cos\left(\frac\pi2+\theta\right)=-\sin\theta$ and $\sin\left(\frac\pi2+\theta\right)=\cos\theta$ with $\theta=\frac{2\pi k}n$:
$$ \boxed{z_k=a-R\sin\frac{2\pi k}n+i\left[b+R\cos\frac{2\pi k}n\right]}, \qquad k=0,1,\ldots,n-1 $$
Verification
At $k=0$: $\ z_0=a-0+i(b+R)=a+(b+R)i$ ✓, matching the given vertex.
| $n$ | Angles of vertices from center (from $z_0$) |
|---|---|
| $3$ | $90^\circ,\ 210^\circ,\ 330^\circ$ |
| $4$ | $90^\circ,\ 180^\circ,\ 270^\circ,\ 0^\circ$ |
| $6$ | $90^\circ,\ 150^\circ,\ 210^\circ,\ 270^\circ,\ 330^\circ,\ 30^\circ$ |
Result
Starting from the "topmost" vertex $z_0$ (directly above the center), successive vertices are reached by rotating counterclockwise by $\frac{2\pi}n$ radians about the center.