Can all conservative vector fields from $\mathbb{R}^2 \to \mathbb{R}^2$ be represented as complex functions? Considering that such a vector field $(M,N)$ is conservative iff for $M,N$ differentiable, $\frac{\partial{N}}{\partial{x}} = \frac{\partial{M}}{\partial{y}}$, we only have one of the two Cauchy-Riemann equations satisfied. Are there then examples of conservative vector fields $\mathbb{R}^2 \to \mathbb{R}^2$ that are not complex-differentiable?
 A: If a scalar function $\phi(x,y)$ on $\mathbb{R}^{2}$ is twice continuously differentiable, then $\nabla \phi = (\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y})$ is a conservative vector field because $\frac{\partial^{2}\phi}{\partial x\partial y} =\frac{\partial^{2}\phi}{\partial y \partial x}$. (That's not much of a restriction on $\phi$, and certainly not enough to guarantee that $\phi$ is the real part of a holomorphic function, for example.)
Conversely, given $(M,N)$ as you describe and continuously differentiable, then you can integrate
$$
       \phi(x,y) = \int_{(0,0)}^{(x,y)}M(x',y')dx' + N(x',y')dy'.
$$
The above is well-defined because the smooth path you choose from $(0,0)$ to $(x,y)$ doesn't matter, which follows from Green's Theorem:
$$
        \oint_{C}Mdx+Ndy = \int\int_{R} \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\,dx dy = 0,
$$
where $C$ is a simple closed positively oriented curve enclosing a region $R$. Because you can choose any smooth path from $(0,0)$ to $(x,y)$, it is not hard to show that $\nabla \phi = (M,N)$.
