$f(z) = z\overline{z}+iv(x,y) = x^2+y^2 + iv(x,y)$ I have to find $v$ so since $f$ is holomorphic it satisfies the Cauchy-Riemann equations :
$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x} = 2x \implies v =2xy + \phi(x) $
$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} = -2y = 2y + \phi'(x)\implies \phi'(x)=-4y \implies \phi(x) = -4xy+C \implies v = 2xy-4xy + C = -2xy + C.\;\;\;\;\;\; C\in \mathbb{C}$
but there is a problem $\frac{\partial u}{\partial x} = 2x \neq \frac{\partial ( -2xy + C)}{\partial y} = -2x$
I'm really confused please can someone shed some light on this contradiction.
thank you.
One can rewire Cauchy-Riemann equations in a single equation $$\dfrac{\partial f}{\partial \bar{z}}=0,$$ by appropriately defining $\dfrac{\partial }{\partial z}$ and $\dfrac{\partial }{\partial \bar{z}}.$ However your function does not satisfies this condition unless $\bar{z}=0$. Therefore it is not holomorphic.