Analytic functions having harmonic real and imaginary parts. I've bee set the following question in a homework assignment for my complex analysis class, but have literally no idea what it means by sufficiently regular.
Let $f : \mathbb{C} \to \mathbb{C}$ be an analytic function, with $f = u + iv$. Prove that if $f$ is sufficiently regular the real
part and the imaginary part of $f$ (i.e. u and v) are harmonic functions in $\mathbb{R}^2$
. That is, if we consider
$u$ as a function of $x$ and $y$ (where $z = x + iy$) we have
$u_{xx} + u_{yy} = 0$     ;    $v_{xx} + v_{yy} = 0$.
Using $u_{xx}$ to denote the 2nd derivative of $u$ with respect to $x$.
We haven't mentioned anything about regularity conditions in lectures and I can't find anything about it in the online notes for the course.
Could someone please point me in the direction as to what it might mean by that? My first thought is that it might be something to do with considering it as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ with $u$ and $v$ as the component functions. But not sure what it means from there.
Thanks in advance
 A: If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations. 
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 v}{\partial y \partial x}$
$\frac{\partial^2 u}{\partial y^2} = - \frac{\partial^2 v}{\partial x \partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$\frac{\partial^2 v}{\partial y \partial x} = \frac{\partial^2 v}{\partial x \partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$.
