Considering how to calculate $n$-th roots of a complex number in polar form, find the vertices of a square of center $0$ knowing that one of them is $(3,4)$.
Ok so I don't know how to express the equation. I know that $w\in\Bbb{C}$ is a $n$-rooth of $z\in\Bbb{C}$ iff $z=w^n$.
In this case we have $n=4$ (because it is a square) and $v_1=3+4i=[5,\arctan(4/3)]$, so Question: the equation becomes $z=v^4$?
Now call $z=[\rho,\varphi]$ and $v=[r,\theta]$. We need to find all $v$ such that $z=v^4$, i.e. $v_k$ for $k\in[1,4]$.
Then $v^4=[r^4,4\theta]$, so by equality of complex numbers in polar coordinates, we have $\rho=r^4$ and $4\theta=\varphi+2k\pi$. Then $v_k=[\sqrt[4]{\rho},(\varphi+2k\pi)/4]$.
Since $v_1=[5,\arctan(4/3)]$ and for the calculated formula $v_1=[\sqrt[4]{\rho},(\varphi+2\pi)/4]$, then $5=\sqrt[4]{\rho}$ and $\arctan(4/3)=(\varphi+2\pi)/4$, hence $\rho=625$ and $\varphi=4\arctan(4/3)-2\pi$.
From here I don't know how to proceed.
Question: How do we calculate $v_2,v_3$ and $v_4$? We could use the polar form of $v_k$ to also find $v_1$ to check if the formula is correct.
EDIT I think my calculations are correct! We end up with $$v_k=\left[5,\frac{4\arctan(4/3)-2\pi+2k\pi}{4}\right]$$ so: $$v_1=[5,\arctan(4/3)]=(3,4)\\ v_2=[5,(4\arctan(4/3)-2\pi+4\pi)/4]=(-4,3)\\ v_3=[5,(4\arctan(4/3)-2\pi+6\pi)/4]=(-3,-4)\\ v_4=[5,(4\arctan(4/3)-2\pi+8\pi)/4]=(4,-3).$$
They are the roots of $z^4=625$, rotated by $\arctan3/4$. The former are $5e^{k\pi i/2},\,k=0,1,2,3$.
Let $w=4+3i$. So, you can use a primitive fourth root of unity, say $e^{πi/2}=i$, to rotate by $\pi/2$ a few times. Get $w,iw,i^2w,i^3w$ as the vertices. That is, $4+3i,-3+4i,-4-3i,3-4i$.
In polar, it's $5e^{i\arctan3/4},5e^{i(\arctan3/4+\pi/2)},5e^{i(\arctan3/4+\pi)},5e^{i(\arctan3/4+3\pi/2)}$.