If $f(z) = u +iv$ is a complex function such that $u$ and $v$ are both harmonic, is $f(z)$ necessarily analytic? I know that all harmonic functions arise as the real parts of analytic; however I am not sure what the correct answer to the question is. I think that if $u$ and $v$ are harmonic then it has to be analytic since an analytic function is one that we can take the derivative of and differentiate. In order to verify if a function is harmonic or not, we have to take the second partial derivative of $u$ and $v$ which can only be done if a function is continuous. Is my thought process correct or do I have any holes/mistakes in my thought process and answer? 
 A: Hint: $u(x,y) = x$ is harmonic. Is $x+ix$ analytic?
A: No, $v$ has to be the harmonic conjugate of $u$, that is $u$ and $v$
have to satisfy the Cauchy-Riemann equations:
$$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$
and
$$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}.$$
If $u$ is harmonic, there's always a solution of these for $v$ locally,
that is in the neighbourhood of any given point.
A: I think you are confusing a few things:
1)If a function is analytic, it can be (locally) written as a power series.
2)A function is harmonic given $\Delta u=0$, where $\Delta$ is the Laplace operator.
3)A complex function is holomorphic (or analytic) if it statisfies the Cauchy-Riemann Eq and is totally differentiable as a real function (from and onto $\mathbb{R}^2$). That implies that the real and imaginary part of $f(z)$ have to be at least differentiable once.  
However, those things imply a few things:
Any real harmonic function is analytic.
The real and imaginary part of a complex, holomorphic (or analytical) function are harmonic.
For a complex valued function, analytic and holomorphic are the same.
