# Cauchy Riemann equation and Harmonic Condition

Question: If harmonic functions $u$ and $v$ satisfy Cauchy-Riemann equations, then $u+iv$ is an analytic function.

Am a bit confused here as we already have a theorem which says that if a function satisfies Cauchy Riemann equations for each point in the domain and each of its four partial derivatives $u_x, v_x, u_y, v_y$ are continuous then it is analytic in that domain.

So can we say that if $u$ and $v$ are harmonic then each of its four partial derivatives $u_x, v_x, u_y, v_y$ are continuous thereby implying that the function is analytic in $D.$

Harmonic functions $u:\mathbb{R}^n\rightarrow \mathbb{R}$ are known to be smooth (i.e. infinitely often continuously differentiable). This is, however, not a completely trivial statement (and the definition of harmonic may be weakened considerably from what is probably known to you).
In most cases, in two dimension, it should be sufficient to know that if a function $u$ is continuously differentiable and you can 'write down $\Delta u$' then the conclusion of the theorem you cited holds true. This is particularly interesting if you have a harmonic function $u$ and want to know whether it's smooth (or even real analytic). The theorem you are using tells you that you can conclude this already if you find a complex analytic extension $u+iv$ which is complex analytic.
Note however that $u, v$ both being harmonic does not imply that $u+iv$ is complex analytic. The $v$ you need for given $u$ is a special one (and locally uniquely determined up to a constant).
• See 5th line of en.wikipedia.org/wiki/Harmonic_function#Remarks Here it says that all harmonic functions are analytic. So, if both $u$ and $v$ are harmonic (implies analytic), then why isn't $u + iv$ analytic? – user265328 Dec 8 '15 at 16:42