Are there closed-form expressions for the real and imaginary functions? By "real and imaginary functions" I mean $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$.
 A: This is cheating, but we can use $\dfrac{z+\bar{z}}{2}$ for the real part, and 
$\dfrac{z-\bar{z}}{2i}$ for the imaginary part.
Here $\bar{z}$ is as usual the complex conjugate of $z$. 
The function that takes $z$ to $\bar{z}$ is not differentiable. That puts some limitations on the kind of auxiliary function one could use for an expression.
Remark: If we don't want to allow complex conjugate as a basic operation, but are comfortable with the norm, then we can define $\bar{z}$ (for $z\ne 0)$, by $\bar{z}=\frac{|z|^2}{z}$.  However, the more natural definition goes the other way, $|z|$ as $\sqrt{z\bar{z}}$.
A: If you look for closed-form expression of $\Re(z) $ as analytic functions such as $\sin(z),e^z,\sqrt {z} $ or combination of known functions or power series etc., it is impossible.
If we prove that $\Re(z) $ cannot be expressed as analytic functions,   $\Im(z)$ cannot either because They have relation $z=\Re(z) +i\Im(z) $.
Lema:  $\Re(z) $ is not analytic function.
Prove: 
$z$ is complex number
$x$ and $y$ are real numbers
$$z=x+iy $$ 
$$\Re(z)=\Re(x+iy)=x$$
 $$\frac{d Re(z)}{d z}=\lim\limits_{ y\to 0  } \frac{d Re(x+iy)}{d (x+iy)}=\frac{\partial (x)}{\partial  x}=1 \tag1$$
$$\frac{d Re(z)}{d z}=\lim\limits_{ x\to 0  } \frac{d Re(x+iy)}{d (x+iy)}=\frac{\partial  (x)}{i \partial  y}=0 \tag2$$
If $\Re{(z)}$ is analytic function Result ($1$) and Result ($2$) must be equal. There cannot be two different limit value on a point.
You can analize all functions in same method if they are analytic or not.
If we want to check 
$f(z)=f(x+iy)=U(x,y)+iV(x,y)$
$$\frac{d f(z)}{d z}=\lim\limits_{ y\to 0  } \frac{d(U(x,y)+iV(x,y))}{d (x+iy)}=\frac{\partial U (x,y)}{ \partial  x}+i\frac{\partial V (x,y)}{ \partial  x} \tag3$$
$$\frac{d f(z)}{d z}=\lim\limits_{ x\to 0  } \frac{d(U(x,y)+iV(x,y))}{d (x+iy)}=\frac{\partial U (x,y)}{ i\partial  y}+i\frac{\partial V (x,y)}{i \partial  y} =-i\frac{\partial U (x,y)}{ \partial  y}+\frac{\partial V (x,y)}{ \partial  y}\tag4$$
If $f(x+iy)$ is analytic function, the result 3 and result 4 must be equal. This means that, the function is complex-differentiable in a whole domain (holomorphic) is the same as an analytic function. 
If we check real and imaginary parts.
We get the result: 
$$\frac{\partial U (x,y)}{ \partial x}=\frac{\partial V (x,y)}{ \partial  y}$$
$$\frac{\partial U (x,y)}{ \partial y}=-\frac{\partial V (x,y)}{ \partial  x}$$
They are named Cauchy–Riemann equations
A: No, closed form should mean something like "easy to calculate" or "easy to manipulate", so $\Re z$ and $\Im z$ are already in closed form. 
