Consider a $\mathbb{R}$-linear function $f:\mathbb{R^2}\longrightarrow \mathbb{C}$, $f(x,y)=(ax+by)+i(cx+dy)$, $a,b,c,d \in \mathbb{R}$. Prove that, identifying $(x,y)$ with $x+iy$, there are $\alpha, \beta \in \mathbb{C}$ such that $f(z)=\alpha z +\beta \overline{z}$, with $ad-bc=|\alpha|^2-|\beta|^2$.
I am puzzled with this question. I have tried to write $\alpha$ and $\beta$ in terms of it real and imaginary parts, and to compare to the expression of $f$, but i got nothing. Any help would be welcome.
Thanks,
Best Answer
One way to do this is to compute both expressions in a way in which you can compare the constants of each expression. So on one hand, plugging in $x + iy$ into $f(z)$ leads to $$f(z) = \alpha(x +iy) + \beta(x - iy) = (\alpha + \beta)x + (i\alpha - i\beta)y.$$ On the other hand, you can take the original expression and write $$f(x, y) = (a + ic)x + (b + id)y.$$
Hence you have a system of equations where $\alpha + \beta = a + ic$ and $i\alpha - i\beta = b + id$. This is just a 2x2 system of equations. You can then solve to find that $\alpha = \dfrac{a + d + ic - id}{2}$ and $\beta = \dfrac{a - d + ic +ib}{2}$. Once you have that, verifying the desired equality $ad -bc = |\alpha|^2 - |\beta|^2$ is straightforward.