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Is there a way to simplify the set of equations\begin{align*} \lvert c_{11}\rvert^2 +\overline{c_{12}}c_{21} & = 1\\ \overline{c_{11}}c_{12} +\overline{c_{12}}c_{22} & = 0\\ \overline{c_{21}}c_{11} +\overline{c_{22}}c_{21} & = 0\\ \overline{c_{21}}c_{12} +\lvert c_{22}\rvert^2 & = 1\end{align*} where the $c_{jk}$ are complex numbers, and the overline indicates complex conjugation?

When I express each $c_{jk}$ as $a_{jk} + ib_{jk}$ where $a_{jk},b_{jk}$ are real, expand the equations, and equate the real and imaginary parts, I get a system of nonlinear equations that I do not know how to handle.

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For convenience of typing, let me write $a=c_{11},b=c_{12},c=c_{21},$ and $d=c_{22}$. So your equations are \begin{align*} \lvert a\rvert^2 +\overline{b}c & = 1\\ \overline{a}b +\overline{b}d & = 0\\ \overline{c}a +\overline{d}c & = 0\\ \overline{c}b +\lvert d\rvert^2 & = 1.\end{align*}

Note that the first equation says in particular that $\overline{b}c$ is real. Assuming $b\neq 0$, this just means that $c=tb$ for some $t\in\mathbb{R}$ (specifically $t=\overline{b}c/|b|^2$). The second equation also allows us to solve for $d$ as $-\overline{a}b/\overline{b}$, and then the third equation will hold automatically since $c$ is a real multiple of $b$. Also, the first equation says that $|a|^2+t|b|^2=1$ so it just uniquely determines $t$ in terms of $a$ and $b$. Finally, $|d|=|a|$ by the formula for $d$ given above and so the fourth equation again just says $|a|^2+t|b|^2=1$.

To sum up, we have found that as long as $b\neq 0$, we can uniquely solve for $c$ and $d$ in terms of $a$ and $b$. Specifically, fix any $a,b\in\mathbb{C}$ with $b$ nonzero. Let $t=\frac{1-|a|^2}{|b|^2}$ and $c=tb$, and let $d=-\overline{a}b/\overline{b}$. Then all four equations will hold, and these are the unique values of $c$ and $d$ that make them hold.

It remains to consider the case where $b=0$. If $b=0$ but $c\neq 0$, the solutions can be described exactly as above just with the roles of $b$ and $c$ swapped (with $b=0$ just meaning that our chosen $a$ must satisfy $|a|=1$). Finally, if $b$ and $c$ are both $0$, the equations just say $|a|=|d|=1$.

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