Determining if any general funtion u(x,y) makes f(z)=u(x,y)+iv(x,y) analytical I have a question about Complex Analytical functions. I have some homework that asks:
let $f(z) = u(x,y) + iv(x,y)$. Indicate the following functions for which u(x,y) may be analytic:
$6(x^2-y^2)$ | $e^{6x}sin(6y)$ | $log(x^2-y^2)$
From what I have seen you need to make sure they satisfy the Cauchy-Riemann equations, so for the first one, $u_x=12x$ and $u_y=12y$. However, to make sure that it's analytical, I need to make sure that $u_x=v_y$ and $u_y=-v_x$. I don't have $v_x$ or $v_y$ unless I deduce them from $u_x$ and $u_y$, but that's proving by assumptions and would make everything Analytic, not for any general function of $v(x,y)$. Am I supposed to do some sort of integrating or differentiating here?
Any help would be appreciated, thanks in advance.
 A: You can in most examples use a short cut: Assuming that $u(x+iy)$ is the real part of some holomorphic function $f(z)$, then
$$
u(x,y)=\tfrac12(f(z)+\overline{f(z)})=\tfrac12(f(z)+\bar f(\bar z))=\tfrac12(f(x+iy)+\bar f(x-iy))
$$
In view of that, if $u(0,0)$ is not singular, then
$$
2u(\tfrac 12z,-\tfrac i2 z)=f(z)+\bar f(0)=f(z)+u(0,0)-iv(0,0)
$$
$v(0,0)$ is a free constant, so one can set 
$$
f(z)=2u\left(\tfrac 12z,-\tfrac i2 z\right)-u(0,0)
$$

If $(0,0)$ is a singularity of $u$, then one can also shift the point that gives the constant, for instance to $(1,0)$,
$$
f(z)=2u(\tfrac12(z+1),-\tfrac i2 (z-1))-u(1,0).
$$

As a first example consider $u(x,y)=6x-4y$. One gets
$$
f(z)=2\left(6\frac z2+4i\frac z2\right)-0=(6+4i)z
$$
and indeed the real part matches the given one.

As an example of the second formula consider $u(x,y)=\ln(x^2+y^2)$. Then
$$
f(z)=2\text{Ln}\left(\tfrac14(z+1)^2-\tfrac14(z-1)^2\right)-\ln(1)
=2\text{Ln}(z)=\ln(|z|^2)+2i\arg(z)
$$
which again reproduces the given real part.

Check your task description, if the application of Cauchy-Riemann was required, then this type of computations can only give some guidance. However, if the task is the completion to a holomorphic function with any means, then one only has to check that this procedure gives indeed the correct real part. Failing that, there is quite possibly no solution, $u$ is not harmonic. Then you should double check that result using the partial derivatives.
A: If $f(z) = u(x,y) + iv(x,y)$ is analytic, then the Cauchy-Riemann equations say that $u_{x} = v_{y}$ and $u_{y} = -v_{x}$.
It follows from this that $u_{xx} = v_{xy}$ and $u_{yy} = -v_{xy}$, hence $u_{xx} + u_{yy} = 0$. A function $u$ which satisfies $u_{xx} + u_{yy} = 0$ is called harmonic, and the reasoning above shows that if $u(x,y)$ is the real part of an analytic function it must be harmonic.
On the other hand, any harmonic function is the real part of an analytic function. To see this, just set $v = \int u_x dy$, and check that the Cauchy-Riemann equations are satisfied.
