Defining $u(x,y)$ in terms of $z$ and $\bar z$ Let $u(x,y)$ be a real valued differentiable function defined on $\mathbb C$.
Then $u$ can also be written as a function of $z$ and $\bar z$.
When explaining this part, my lecturer jumped straight to claiming that 
$$x = \frac{1}{2}(z + \bar z)$$
$$y = \frac{1}{2i}(z - \bar z)$$
I've been at this for awhile now and i can't seem to figure out why this is so.
This is my thoughts on it: The value of $x$ is the real part of $u$ which would mean that performing $z+\bar z$ effectively nullifies the imaginary terms of $z$ leaving $2 \times Re(z)$, which is why the value was halved.
As for $y$, $z-\bar z$ cancels out the real terms leaving only $2\times Im(z) \times i$, which is why it was further divided by $2i$.
So now we have $u(x,y) = Re(z) + i\times Im(z)$? I don't see why this is the form described above. Is this the only real valued differentiable function for $u$? If not, why was it ok to make this assumption?


So now we have $u(x,y) = Re(z) + i\times Im(z)$?

Not quite.  We have $x+iy = \mathop{\mathrm{Re}}(z) + i \mathop{\mathrm{Im}}(z)$.  So we can write a function $\tilde u(z,\bar z)$ as
$$
   \tilde u(z,\bar z) = u\left(\frac{1}{2}(z+\bar z),\frac{1}{2i}(z-\bar z)\right)
$$
Through a slight abuse of language we consider $u$ and $\tilde u$ to be the same function.
