Given u and v are harmonic in some region R prove the following If u and v are harmonic in some region R prove the following 
$ ( \frac {\partial u}{ \partial y} - \frac {\partial v}{ \partial x}) + i(\frac {\partial u}{ \partial x} + \frac {\partial v}{ \partial y})$
is analytic in R. Does this mean if u and v are harmonic in R then their first order partials in R satisfy the Cauchy-Riemann equations and if so how do I prove it? I tried taking the partial derivatives of each term to see if they would satisfy the Cauchy-Riemann equations but that didn't work.

Define the function $\hat{u}(x,y) := \frac{\partial u}{\partial y}(x,y) - \frac{\partial v}{\partial x}(x,y)$ and $\hat{v}(x,y) := \frac{\partial u}{\partial x}(x,y) + \frac{\partial v}{\partial y}(x,y)$. Then consider the function $f(x,y) := \hat{u}(x,y) + i \cdot \hat{v}(x,y)$, this is your function in question. Now what do you have to check if you want $f$ to be analytic in the region $R$? Of course, you need to check wheter the Cauchy-Riemann equations are satisfied or not. Consider: $\hat{u}_x = \frac{\partial^2 u}{\partial y \partial x} - \frac{\partial^2 v}{\partial x^2}$ and $\hat{v}_y = \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 v}{\partial y^2} \Longrightarrow \hat{u}_x - \hat{v}_y = \frac{\partial^2 u}{\partial y \partial x} - \frac{\partial^2 v}{\partial x^2} - \frac{\partial^2 u}{\partial x \partial y} - \frac{\partial^2 v}{\partial y^2} = 0$, since $v$ is harmonic and $u$ has continous partial derivatives, hence the order of differentiation doesn't matter (by Schwarz's theorem), so the mixed terms cancel out each other. This shows that your function $f$ satisfies the first part of the CR equations. Now, in the same manner, check the second part of the CR equations, and you'll arive at the desired result.
