@Mark Yes. I hope so.

@KhashF Do you mean that if $A(z) $ (with entire entries) is a normal matrix on the real line it will be normal on the whole complex plane?

$A^{*}$ is not the same as $A^{\#} $. $A^{\#} (z) =A^{*} (\bar{ z}) $.

Ok. I only have a question about the first line when you defined $v$: I know that $v(x, y)=\Im(f(z)) =\frac{f(z) - \overline {f(z)}} {2i}$. So, is your $v$ is the same as the imaginary part of $f$? According to what you wrote it is like having $v(x, y)=\frac{f(z) - \overline {f(\bar{z} )}} {2i}$!!!

the content $b$ depends on the choice of the function $g$, is this wright?

thank you, this is true, and contains more. The function $f(z) =-iz=-ix+y$ satisfy the condition above but it is not real for real $z$.