Let $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. I want to use the Cauchy Riemann equations to find all differentiable functions of the form
$$Re( h(z))=2x^2+2x+1-2y^2$$
So I used the C-R equations with $u(x,y)=2x^2+2x+1-2y^2$:
$$\frac{\partial u}{\partial x}= 4x+2=\frac{\partial v}{\partial y} \quad \quad \implies v=4xy+2y+h(y)$$
$$\frac{\partial u}{\partial y}= -4y=- \frac{\partial v}{\partial x} \quad \quad \implies v=4xy+g(x)$$
But I don't really know what this means for the value of $v$. Obviously both terms match up with the $4xy$, but then what about the extra terms?
So thus far you have $$v=4xy+2y+h(y)=4xy+g(x).$$ This means $$2y+h(y)=g(x).$$ Left side is a function of $y$ (only) and right side is that of $x$ (only) but the two have to be equal. This can only happen if both of them are constant functions. So \begin{align*} g(x) & = c\\ h(y) & = -2y+c \end{align*} Thus $v=4xy+c$.