Suppose f(x,y,z) maps $\mathbb{R}^3\rightarrow\mathbb{C}^1$. That is, it takes in three real numbers and spits out a complex number. Does the following always hold:
$$\vec\nabla \text{Re}[f(x,y,z)]=\text{Re}[\vec\nabla f(x,y,z)],$$ where $\vec\nabla$ denotes the gradient and "$\text{Re}$" means take the real part. It seems to be okay for some specific cases, but is it true in general? If so, how can you show it?
There exist real-valued functions $g:\mathbb R^3\to \mathbb R$ and $h:\mathbb R^3\to \mathbb R$ such that $f(x,y,z) = g(x,y,z) + ih(x,y,z)$. If $g$ and $h$ have all three partial derivatives, then we have $\vec\nabla f = (f_x,f_y,f_z) = (g_x+ih_x,g_y+ih_y,g_z+ih_z) = \vec\nabla g + i\vec\nabla h$.
Hence $\vec\nabla \Re(f) = \vec\nabla g = (g_x,g_y,g_z) = \Re \vec\nabla f$.
The question remains: do $g$ and $h$ have all three partial derivatives? To keep it simple, let $F:\mathbb R\to \mathbb C$ be differentiable, with real and imaginary parts $F = G + iH$. Are $G$ and $H$ necessarily differentiable? Yes they are, because
$$\frac{G(x+h)-G(h)}{h} = \Re\left(\frac{F(x+h)-F(h)}{h}\right)$$
(similarly for $H$), and the right-hand side tends to a limit as $h \to 0$ because $\dfrac{F(x+h)-F(h)}{h}$ does.