I have a problem somewhere ... I missed a point..
So given are following equations: (with $C \to R$)
(i) $a(z) = x^2 - y^2$
(ii) $b(z) = x^2 + y^2$
And I am supposed to find out if those equations could be the real component of an complex equation ($C \to C$) I used the Cauchy Riemann equation which gave me
- (i) $2x=0$ and $-2y=0$
- (ii) $2x=0$ and $2y=0$
Both just work for $x = y = 0$
But this would mean $z = 0$ right? So neither of them can be a component of an complex equation (as in complex differentiable function) right?
Based on the information you gave in the comments section, i assume that the exercise’s question migth be rewritten as:
Figure out if the real-valued functions $\{a;b\}:\mathbb{C\rightarrow R}$, given by: \begin{array} \\a(z)=x^2-y^2 \\b(z)=x^2+y^2 \end{array} with $z=x+yi$, can be the real part $\Re(f)$ of a complex function $f:\mathbb{C\rightarrow C}$.
With these premises, the exercise has a defined answer: Let $f_a$ and $f_b$ be complex functions. If we let: $$f_a(z)=z^2=x^2-y^2+2xyi$$ We see that $\Re(f_a)=a$, so $f_a$ exists. Likewise, if we let: $$f_b(z)=|z|=z\overline{z}=x^2+y^2$$ Wich is it’s own real part, so $f_b$ exists too. In other words both $a$ and $b$ can be the real part of a complex function $f:\mathbb{C\rightarrow C}$.
As for the cauchy-riemann equations, you’re using them the wrong way. For the equations to work, you need the initial function $f(z)$ to be able to decompose to the form: $$f(z)=u(x;y)+iv(x;y)$$