# How to simplify conditions for conjugations on $\Bbb C^2$

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.

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 $$|a|=|d|=1$$.