# 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))

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I'm confused. If $\log(z)$ gives you the angle, then you must be able to find the imaginary part of $\log(z)$; but then why can't you find the imaginary part of $z$ to begin with? Note that $\log(z)=\log(r)+i\theta$, where $r$ is the modulus $z$, $\log(r)$ is the real log of $r$, and $\theta$ is an argument for $z$. –  Jonas Meyer Mar 23 '11 at 1:15
I'm not sure what you're looking for, but there is no way to express $\overline z$ as an elementary function of $z$ (in the sense of en.wikipedia.org/wiki/Elementary_function), because elementary functions are holomorphic, whereas $z\mapsto\overline z$ is not. For what it's worth, you can find $\overline z$ if you can find $|z|$, because $\overline z = |z|^2/z$. –  Jonas Meyer Mar 23 '11 at 1:23
You can express conjugation as an arithmetic function in the quaternions, but not in the complex numbers. –  Alex Becker Mar 23 '11 at 1:57
@Alex: any justification why? That might be good as a full answer. –  Mitch Mar 23 '11 at 14:00
The point made by @Jonas Meyer is that conjugation does not satisfy the Cauchy–Riemann equations and so is not holomorphic. In particular, it's not an elementary function (ie, does not have a close-form expression). –  lhf Mar 23 '11 at 14:39

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.

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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 $y$.

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Yes, if you know one pair then you know the other. The point is that if you know neither pair, $(x,y)$ or $(z, \bar{z})$, then can you determine $\bar{z}$ from $z$? (and the answer is presumably 'no'. –  Mitch Mar 24 '11 at 13:40

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.

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I am processing the criteria for 'holomorphic', and it makes sense via symbols, but I'm having trouble translating that to a good intuitive feel for it (especially since visualizing a complex function is not so immediate, essentially a 4-d 'thing' to visualize (or 3-d with colors)). Any hints there? (it may be obvious, but I don't have much experience with complex (or real for that matter) analysis). –  Mitch Mar 24 '11 at 19:35
@Mitch: Yeah, the "graph" of a complex function would be 4 dimensional, so you can't visualize it directly. Instead you can try picturing how the functions transform portions of the plane. If you zoom in close enough, holomorphic functions look like rotations and dilations. If $f$ is complex differentiable at $a$, then $f(z)-f(a)\approx f'(a)(z-a)$ when $z$ is close to $a$. This says that if you zoom in on $a$, the behavior of $f$ is described by multiplication by the complex number $f'(a)$. The modulus of $f'(a)$ represents a dilation/compression, and the argument represents a rotation. –  Jonas Meyer Mar 25 '11 at 1:57
I'm still having trouble with visualizing even this...what does the derivative of the 'conjugate' function look like? (or I'm guessing that maybe it's not possible at any point? I just don't 'see' any of this.) –  Mitch Mar 26 '11 at 13:57
@Mitch: You can think of the derivative at a point as being a linear transformation of the plane. If $f'(a)$ exists, then the corresponding linear transformation is $w\mapsto f'(a)w$. If $f'(a)=re^{i\theta}$, then $f'(a)w=r|w|e^{i(\arg{w}+\theta)}$, which corresponds to rotation of the plane by $\theta$ and "rescaling" by $r$. Now, complex conjugation cannot correspond to such a transformation, because it sends $w$ to $|w|e^{-i\arg w}$. Conjugation is real linear, and real differentiable; but since it (along with its derivative) is not complex linear, it is not complex differentiable. –  Jonas Meyer Mar 27 '11 at 4:34
@Mitch: A function that is linear when $\mathbb{C}$ is thought of a real vector space but not when it is thought of as a complex vector space is not complex differentiable. Identifying $\mathbb{C}$ with $\mathbb{R}^2$, the linear maps can be described with 2-by-2 matrices of real numbers. The dilation/rotations correspond to matrices of the form $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$; this also corresponds to multiplication by the complex number $a+bi$. Conjugation has matrix $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, which does not correspond to a complex number, and is not complex linear. –  Jonas Meyer Mar 27 '11 at 4:43