Find area of the polygon with corners defined by the roots of $\sqrt{7}+3i-x^{2n}=0$, as $n\to \infty$.
Source: IME 1967 (Instituto Militar de Engenharia, Brazil, Entrance Exame, no answer provided)
My solution:
The polygon is defined by the $2n$ roots of $x^{2n}=\sqrt{7}+3i=4(\frac{\sqrt{7}}{4}+i\frac{3}{4})$, defined by $$x_k=4^{\frac{1}{2n}}\text{cis}(\frac{\theta + 2k\pi}{2n}), k=0,\ldots,2n-1.$$
The polygon is regular with $2n$ corners and converges to a circle centered at (0,0) with radius $2^{\frac{1}{n}}$, as $n\to \infty$ with area $$\lim_{n\to \infty} A_n=\pi\ 2^{\frac{2}{n}}=\pi$$
Question/Comment: I believe the solution is correct but is rather informal/based on intuition. I would like a more formal answer, based on algebraic arguments. Is it possible?
Help is appreciated. Sorry if this is a duplicate.
Given that $x_k = 2^{\frac1n}e^{i\frac{\theta+2k\pi}{2n}}$, there are $2n%$ isosceles triangles with side length $a=2^{\frac1n}$ and vertex angle $\frac{\pi}n$. Thus, the area of the $2n$-polygon is
$$A_n = 2n\cdot \frac12 a^2\sin\frac{\pi}n =2^{\frac2n} n\sin\frac\pi n $$
After take the limit, the area approaches
$$\lim_{n\to\infty} A_n = \lim_{n\to\infty} 2^{\frac2n} \cdot \lim_{n\to\infty} \frac{\sin\frac\pi n}{\frac1n} = 1\cdot \pi =\pi$$