Complex coordinates of the vertices of a polygon

Question

If $z_0$ be the centre of a regular polygon of $n$ sides and $z$ be its one vertex $A_1$, then the vertices $A_2, A_3,\dots, A_n$ (proceeding in anticlockwise direction, taking actual position of points) will be: $z_0+(z-z_0)\alpha, z_0+(z-z_0)\alpha^2,\dots, z_0+(z-z_0)\alpha^{n-1}$ respectively where $\alpha=e^{\frac{\iota 2\pi}{n}}$. If the points are taken in clcokwise direction then $\alpha$ must be taken as $e^{\frac{-\iota 2\pi}{n}}$

I read this in a book. There was no proof given for this and I cannot understand it. Please help me understand this.

Answer 1

If you would make a sketch of this for yourself, you would see that $z-z_0$ can be thought of as a vector pointing outward from $z_0$. If you multiply that by $e^{\frac{i2\pi}{n}}$ that means you are rotating that vector by the angle $\frac{i2\pi}{n}$, relative to the origin, therefore you must add $z_0$ to that to place it properly. This step is repeated $n-1$ times. Bear in mind that each time you say $(e^{\frac{i2\pi}{n}})^m$ you are merely increasing the rotation angle m-fold.

Answer 2

The centre of the regular polygon is represented by $z_0$ and its vertices by $z_1, z_2, \dots, z_n$ (where $n$ is the number of sides) in anticlockwise order. The anticlockwise angle formed by joining $z_1, z_0$ and $z_k$, where $k\in \{2, 3,\dots,n\}$, is equal to $\frac{2\pi(k-1)}{n}$. Also, the sides of a regular polygon are equal.

Hence, $\frac{z_k-z_0}{z_1-z_0}=\frac{|z_k-z_0|}{|z_1-z_0|}e^{i\frac{2\pi(k-1)}{n}}=e^{i\frac{2\pi(k-1)}{n}}=\alpha^{k-1}$

So, $z_k-z_0=(z_1-z_0)\alpha^{k-1}$

$\therefore z_k=z_0+(z_1-z_0)\alpha^{k-1}$

Putting $k=2,\dots,n-1$ in the above equation, we obtain the complex coordinates of all the remaining vertices of the polygon.

We obtain a similar result, when the angle is measured in clockwise sense using the same approach.