Formula for calculating moduli of roots of complex numbers from determinants - not working Determinants are used here as scaling factors for the areas of the polygons generated by connecting the individual roots of $z$.


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*The roots of complex numbers are given by the formula $$z_k=\sqrt[n]{r}\Big(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\Big)$$

*Two consecutive roots ($k=0$ and $k=1$) can be represented as
$$\begin{bmatrix}\sqrt[n]{r}(\cos\frac{\theta}{n})&\sqrt[n]{r}(\cos\frac{\theta+2\pi}{n}) \\ \sqrt[n]{r}(\sin\frac{\theta}{n})&\sqrt[n]{r}(\sin\frac{\theta+2\pi}{n})\end{bmatrix}$$

*And the determinant of that, after some algebraic manipulation, is given by $$\text{det}=\sqrt[n]{r^2}\Big(\sin\frac{2\pi}{n}\Big)$$

*After excluding the cases where $\displaystyle\sin\frac{2\pi}{n}$ is $0$, I arrived with a final formula of
$$r=\Big(\frac{\text{det}}{\sin\frac{2\pi}{n}}\Big)^{\frac{n}{2}}$$


However, the values generated by this formula are incorrect. Here is an example with $n=4$, $\text{det}=3$; the resultant area is $162$ instead of $6$).
The area $a$ for a polygon with $n=4$ and $r=1$ is $4(\frac{r^2}{2})=4(\frac{1}{2})=2$.
The area $A$ for the same polygon is then expected to be $a\times\text{det}=2\times3=6$. Instead I get the following:
$$r=\Big(\frac{\text{det}}{\sin\frac{2\pi}{n}}\Big)^{\frac{n}{2}}=\Big(\frac{3}{\sin\frac{2\pi}{4}}\Big)^{\frac{4}{2}}=9$$
$$A=4(\frac{r^2}{2})=4(\frac{9^2}{2})=162\neq6$$
At which step did I make an error? Are my assumptions correct in the first place?
Edit: My biggest concern right now is whether complex numbers are allowed to be represented as in step 2.
 A: Let me write some words, from your statements. Maybe it works:
Suppose all vertices of a regular polygon are on a circle with radius $r$.
(1). The n-th roots of 1 as a complex number are given by the formula $$z_k=\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}$$
(2). All of these roots lie on circle $|z|=r$, so following points are the vertices of regular polygon 
$$w_k=r\Big(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\Big)$$
(3). Every two consecutive points (points $k$ and $k+1$) with origin as triangle third vertex, swap an area $A_k$ equal to
$$A_k=\frac12\text{det}\begin{bmatrix}r\cos\frac{\theta+2k\pi}{n}&r\cos\frac{\theta+2(k+1)\pi}{n}\\r\sin\frac{\theta+2k\pi}{n}&r\sin\frac{\theta+2(k+1)\pi}{n}\end{bmatrix}$$
(4). And the determinant of that, after some algebraic manipulation, gives us
$$A_k=\frac12r^2\sin\frac{2\pi}{n}$$
(5). After adding all areas, the total area of polygon is
$$A=\sum_{k=0}^{n-1}A_k=\frac{nr^2}{2}\sin\frac{2\pi}{n}$$
(6). If our process were done in true way, We must have
$$\lim_{n\to\infty}A=\pi r^2$$
is the area o circle with radius $r$, because as $n\to\infty$, the polygon will convert to circle.
For example with $n=4$, the area of square is
$$A=\frac{4r^2}{2}\sin\frac{2\pi}{4}=2r^2$$
With $n=6$, the area is
$$A=\frac{6r^2}{2}\sin\frac{2\pi}{6}=\frac{3\sqrt{3}}{2}r^2$$

