Complex conjugate of $z$ I am just starting to learn complex analysis, and it appears that you can have a function $f = u+iv$, and thus you can 
have $$f(z) = u(x,y)+iv(x,y).$$
Am I incorrect in believing this?
If this is true, then what do $u(x,y)$ and $v(x,y)$ mean?
I believe $u$ and $v$ are just functions of the real and imaginary parts of $z$, why not just have $u(x)$ and $v(y)$ instead?
Also, does this mean that
$$f(\bar z) = u(x,y)-iv(x,y)?$$
 A: The idea is that you can look at the real and the imaginary part of the function. If $f:\mathbb{C}\rightarrow \mathbb{C}$, $f(z)=u(z)+iv(z)$, where $u(z)=\text{Real part of }f(z)$ and $v(z)=\text{Imaginary part of }f(z)$. $u,v:\mathbb{C}\rightarrow \mathbb{R}$. It is easier to study functions that have real values.
Considering the writing of $z=x+iy$, $x,y\in\mathbb{R}$, you can identify $\mathbb{C}$ with $\mathbb{R}$ and consider $u,v:\mathbb{R}^2\rightarrow \mathbb{R}$ as above.
Since $f$ has argument $z$ you cannot define $u$ only using $x$. For example $f(z)=z^2$.
$f(x+iy)=x^2-y^2+2xyi$. $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$, both depending on both $x$ and $y$.
A: $u(x,y)$ is the real part of $f(x,y)$ and $v(x,y)$ is the imaginary part of $f(x,y)$.
Whilst it is true that $\overline{f(z)} = u(x,y)-iv(x,y)$, it is not necessarily true that ${f(\bar z)} = u(x,y)-iv(x,y)$
Here's an example where ${f(\bar z)} = u(x,y)-iv(x,y)$
Let $f(z)=z^2$
Then $f(x+iy)=(x+iy)^2= x^2-y^2+2ixy=(x^2+y^2)+i(2xy)$
Rewrite as $f(x,y)=(x^2+y^2)+i(2xy)=u(x,y)+iv(x,y)$
where $u(x,y)= x^2+y^2$ and $v(x,y)= 2xy$
$\overline {f(z)} = (x^2+y^2)-i(2xy)$
${f(\bar z)} = f(x-iy)=(x-iy)^2=x^2+y^2-2ixy=(x^2+y^2)-i(2xy)= \overline {f(z)}$
BUT
Here's an example where ${f(\bar z)} \ne u(x,y)-iv(x,y)$
Let $f(z)=iz$
Then $f(x+iy)=i(x+iy)= -y+ix=(-y)+i(x)$
Rewrite as $f(x,y)=(-y)+i(x)=u(x,y)+iv(x,y)$
where $u(x,y)= -y$ and $v(x,y)= x$
$\overline {f(z)} = (-y)-i(x)$
${f(\bar z)} = f(x-iy)=(x-iy)i=xi +y=(y)+i(x) \ne \overline {f(z)}$
A: In the same way as we write $z=x+iy$ we can write $f(z) = u(x,y)+iv(x,y):$ $u$ is the real part of $f(z)$ and $v(x,y)$ is the imaginary part. 
On the other hand, it is $f(\bar{z}) = u(x,-y)+iv(x,-y)$ which, in general, is different from $u(x,y)-iv(x,y).$ 
Just consider $f(z)=z.$ It is $$f(z)=z+i=x+i(y+1).$$ Thus, $u(x,y)=x$ and $v(x,y)=y+1.$ Note that $$f(\bar{z})=x-iy+i\ne u(x,y)-iv(x,y)=x-i(y+1).$$
A: Every complex number $z$ can be thought of as an ordered pair:
$$z = x+iy = (x,y),$$
and so any function $f:\mathbb{C}\to\mathbb{C}$ can be thought of as a function $f:\mathbb{R}^2\to\mathbb{R}^2$. Since every complex number has a real and imaginary part, every output of $f:\mathbb{C}\to\mathbb{C}$ will have a real and imaginary part. In your first equation, $u$ is the real part of the function and $v$ is the imaginary part. Just keep in mind that 
$$\overline{f}(z) = u(x,y)-iv(x,y) \neq f(\bar{z})$$
in general, and the others have given examples of where the equality you have fails.
A: When we write "let $z=x+iy\>$", or "let $f(z):=u(x,y)+i v(x,y)\>$" and similar things there is some sloppy notation involved. We can straighten matters out by explicitly introducing the map
$${\rm cpl}:\quad {\mathbb R}^2\to{\mathbb C}, \qquad(x,y)\mapsto x+iy$$
and its ${\mathbb R}^2$-valued inverse
$${\bf re}:\quad {\mathbb C}\to{\mathbb R}^2,\qquad z\mapsto\bigl({\rm Re}(z), \>{\rm Im}(z)\bigr)\ .$$
One would then have $z={\rm cpl}(x,y)$ and
$$f(z):={\rm cpl}\bigl(u({\bf re}(z)),\>v({\bf re}(z))\bigr)\ .$$
One could even condense the functions $(x,y)\mapsto u(x,y)$ and $(x,y)\mapsto v(x,y)$ into $${\bf w}:\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad (x,y)\mapsto {\bf w}(x,y):=\bigl(u(x,y),v(x,y)\bigr)$$ and write
$$f={\rm cpl}\circ{\bf w}\circ{\bf re}\ .$$
But in real life dealings with such matters all of this is "tacitly understood", and after getting accustomed to it one considers musements of the above kind as "generalized abstract nonsense".
A: Functions like
$$f(z)=u(x)+iv(y)$$ are a special case of $$f(z)=u(x,y)+iv(x,y).$$
They are less interesting than the general case because they are "separable": taking the derivative, one of the arguments just vanishes:
$$\frac{\partial f(z)}{\partial x}=u'(x),\\\frac{\partial f(z)}{\partial y}=i^2v'(y).$$
This makes them similar to two independent functions.
A deeper reason is that such functions do not fulfill the Cauchy-Riemann equations and are not derivable with respect to $z$. As you'll learn later, this makes them less interesting mathematical objects, as differentiability on $z$ has numerous wonderful implications.
