Complex conjugate of $z$ without knowing $z=x+i y$ Is it possible to determine (and if so, how) the complex conjugate $\bar{z}$ of $z$, if you don't already know that $z = x + i y$? 
I think you can use $\log(z)$ to get the angle, and therefore the ratio of $y$ and $x$. But how do you get $|z|$, the radius? How then to get $r$ (so that $x = {\rm Re}(z) = r \cos(\log(z))$ and $y = {\rm Im}(z) = r \sin(\log(z))$)? 
(this is related to How to express in closed form?, namely you can compute Re and Im using the conjugate, but then how do you reduce the conjugate itself fully to elementary functions (if at all))
 A: As others have pointed out, since $\bar{z}$ is not holomorphic, there is no elementary function representation in terms of $z$.
The intuitive reason that complex conjugation is not holomorphic is that it reverses orientation.
More precisely, any holomorphic function is locally conformal (since the zeroes of its derivative are isolated). But conformal means, among other things, that orientation is preserved.
Since complex conjugation reverses the orientation of the entire complex plane, it can't be conformal, and thus it is not holomorphic.
A: If you knew $z$ and $\bar{z}$ then you could easily recover the real and imaginary parts $x$ and $y$, since:
$x = \frac{z+\bar{z}}{2}$
$y = \frac{z-\bar{z}}{2i}$
So you certainly can't know $z$ and $\bar{z}$ without also knowing $x$ and $y$.
A: After reading your last comment, I thought it might help to elaborate on my comment in an answer.
The elementary functions are holomorphic because exponentials and powers and their inverses are holomorphic, constants are holomorphic, arithmetic combinations (sums, products, etc.) of holomorphic functions are holomorphic, and compositions of holomorphic functions are holomorphic.
The function $f(z)=\overline{z}$ is not holomorphic because its complex derivative does not exist at any point.  If $z\neq w$, then $\displaystyle{\frac{f(z)-f(w)}{z-w}=\frac{\overline{z-w}}{z-w}=\exp(-2i\mathrm{arg}(z-w))}$.  You can have $z$ arbitrarily close to $w$ while $\exp(-2i\mathrm{arg}(z-w))$ can be an arbitrary point on the unit circle.  To see this, you can consider $z$ moving in tiny circles around $w$.  To see more explicitly what can go wrong, consider $z$ approaching $w$ along the horizontal line $z=w+t$, where $t$ is real and approaches zero, and then consider $z$ approaching $w$ along the vertical line $z=w+it$, where again $t$ is real and approaches $0$.  In the first case the quotient is always $1$, and in the second case the quotient is always $-1$.  Therefore $\displaystyle{\lim_{z\to w}\frac{f(z)-f(w)}{z-w}}$ does not exist.  
The Cauchy-Riemann equations give a less direct (but perhaps easier) way to see that $f$ is not holomorphic.  If $g$ is a holomorphic function, and if $g$ is expressed as a function of $x$ and $y$ (with a little abuse of notation) $g(x,y)=g(x+iy)$, then $\displaystyle{i\frac{\partial g}{\partial x}=\frac{\partial g}{\partial y}}$.  In the case of $f(z)=\overline{z}$, you have $f(x,y)=x-iy$, so $\displaystyle{i\frac{\partial f}{\partial x}=i}$ and $\displaystyle{\frac{\partial f}{\partial y}=-i}$, showing that the Cauchy-Riemann equations are not satisfied.
