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I am teaching complex analysis to physics students next semester and I would like to discuss the fact that $z$ and $\bar{z}$ are functionally(?)/algebraically(?) (what is the correct terminology, anyway?) independent as early as possible in a course. Ideally, after I introduce complex numbers $z$ as a pair of real numbers and develop the basic complex algebra, I would like to define the complex conjugate $\bar{z}$ and then prove that $\bar{z}$ cannot be obtained from $z$ using algebra, i.e. using $+,-,\times,\div$. Now, my question is, what is the strongest statement of independence of $z$ and $\bar{z}$ that I can prove using basic complex algebra?

Rephrasing the question: how can I prove that $z$ and $\bar{z}$ are algebraically independent in the ordinary sense of the word? I want to prove there exists no nontrivial polynomial $P_n(z,\bar{z})$ with complex constants so that $P_n(z,\bar{z}) = 0$ for any $n \in \mathbb{N}$ (including $n \to \infty$).

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    in general the notation for complex conjugate is $\bar z$. What is you definition of independence, as linearly independent? –  Jul 09 '17 at 20:39
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    Actually $\bar{z}=\dfrac{|z|^2}{z}$ for any $z\in\mathbb{C}^*=\mathbb{C}\setminus {0}.$ – mfl Jul 09 '17 at 20:41
  • @Masacroso My definition of independence is definitely broader than linear independence. I want to stress that you cannot get $\bar{z}$ using algebraic manipulations on $z$. –  Jul 09 '17 at 20:43
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    @mfl I would rather define the norm later on as $|z|^2 = z \bar{z}$, –  Jul 09 '17 at 20:44
  • Why the downvotes? Is the question not clear? –  Jul 09 '17 at 20:46
  • @LBO you must define better what are you searching for... Algebraic manipulations of what kind... of $z$ over itself? You stated the operations $+,-,\times,\div$... but these operations are binary. –  Jul 09 '17 at 20:46
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    I'm guessing the OP has the well-known question about how to make sense of $$\frac{\partial}{\partial\overline{z}}\quad\text{and}\quad\frac{\partial}{\partial z}$$ – GEdgar Jul 09 '17 at 20:50
  • @Masacroso I guess I want to say that there is no analytic $f(z)$ so that $\bar{z} = f(z)$. –  Jul 09 '17 at 20:50
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    It is clear that the function $f(z)=\overline z$ is not analytic. Indeed that function coincides with the identity on the real line, hence would have to coincide with the identity everywhere. But I doubt that this is what you meant by "independent". – lulu Jul 09 '17 at 20:57
  • @GEdgar Yes, something along the lines of that. I want to develop an intuitive notion of the independence of $z$ and $\bar{z}$ early on in the course, even before I touch upon differentiability. –  Jul 09 '17 at 20:58
  • @LBO : never mind the downvotes or the votes to close, good question, upvote from me. I dont understand what is their beef with this question. – jimjim Jul 09 '17 at 21:21
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    @LBO (including n→∞) Not entirely clear what that means. Leaving the infinite case aside, for finite $n$ you can write $,P(z, \bar z)=P(x+iy, x-iy)=A(x,y)+iB(x,y),$ where $A,B$ are real polynomials. Then use the result from Can nonzero polynomials vanish identically? to show that $A=B\equiv 0,$. – dxiv Jul 09 '17 at 23:39
  • @dxiv The $n \to \infty$ case should include not only algebraic independence, but also a sort of "transcendental independence". –  Jul 09 '17 at 23:45
  • I've generalized the result to power series (basically is the same) if this is what you want. I cannot find any suitable definition of "trascendental independence". – Overflowian Jul 10 '17 at 12:56

3 Answers3

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Maybe this is what you are searching for: consider the complex cotangent space of $\mathbb{C} $ at $0$ denoted as $T^*_0(\mathbb{C})^\mathbb{C} = Hom_\mathbb{R} (\mathbb{C},\mathbb{C})$. It is a complex vector space of dimension 2, indeed identifying $\mathbb{C}$ with $\mathbb{R}^2$ gives that the forms $dx $ and $dy$ generate (complex linear combinations) the whole $T^*_0(\mathbb{C})^\mathbb{C}$. Now if you put $dz = dx + i dy $ and $d\overline{z} = dx - i dy$ you obtain other two generators, in particular $dz$ and $d\overline{z}$ are linearly independent as element of the complex vector space $Hom_\mathbb{R} (\mathbb{C},\mathbb{C})$.

(after the edit) From what said above it follows that given $f\in C^1(\mathbb{C},\mathbb{C} )$, we have that $df = \frac {\partial f} {\partial x} dx + \frac {\partial f} {\partial y} dy = \frac {\partial f} {\partial z} dz + \frac {\partial f} {\partial \overline{z}} d\overline{z} $ (identifying the differential as a section of $T^*(\mathbb{C})^\mathbb{C}$). Let $P(x_1,x_2)\in \mathbb{C}[x_1,x_2]$ be a complex polynomial such that $ \forall z \in \mathbb{C} \ P(z,\overline{z}) = 0$. Then $0 = dP = \frac {\partial P} {\partial z} dz + \frac {\partial P} {\partial \overline{z}} d\overline{z} $. But $\frac {\partial P} {\partial z} (z,\overline{z}) = \frac {\partial P} {\partial x_1} (z,\overline{z})$ and $\frac {\partial P} {\partial \overline{z}} (z,\overline{z}) = \frac {\partial P} {\partial x_2}(z,\overline{z})$ which implies that $P(x_1,x_2)$ is constant as a polynomial (and must be zero by the hypothesis). This can be generalized also to $f \in \mathbb{C}\{x_1,x_2\}$ (the $\mathbb{C}$-algebra of convergent power series in a neighborhood of $0\in \mathbb{C}^2$). The proof is the same since the convergence of the series is uniform on compact sets (so we can exchange limit and derivative), therefore it holds again that $\frac {\partial f} {\partial z} (z,\overline{z}) = \frac {\partial f} {\partial x_1} (z,\overline{z})$ and $\frac {\partial f} {\partial \overline{z}} (z,\overline{z}) = \frac {\partial f} {\partial x_2}(z,\overline{z})$.

Overflowian
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Well, I don't know of the strongest result, but one can show that if $f(x,y)=u(x,y)+iv(x,y)$ is analytic, then: $\frac{\partial f}{\partial \bar{z}}=0$.

First note that $x=\frac{z+\bar{z}}{2}$ and $y=\frac{z-\bar{z}}{2i}$

Then: $\frac{\partial f}{\partial \bar{z}}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \bar{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar{z}}= (\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x})\frac{1}{2}+(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y})\frac{-1}{2i}= \frac{1}{2}((\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})+i(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}))$.

That by Cauchy-Riemann vanishes.

Masso
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The difficulties arise because we use the same letter $z$ to denote some individual complex number (and then $\bar z$ is the complex conjugate of $z$), and (b) as name for the identity function on ${\mathbb C}$. In the second interpretation $\bar z$ is some other complexvalued function on ${\mathbb C}$. You now ask whether this other function might be analytically related to the function $z$. In other words: Is there an analytic function $f$ (polynomial, algebraic expression, or more complicated), defined at least in some neighborhood $U$ of the origin, such that $$\bar z =f(z)\qquad(z\in U)\quad ?\tag{1}$$ In order to prove that this is not the case we make use of the underlying real structure. From $\bar z=x-iy$ it follows that $${\partial\bar z\over\partial x}=1, \qquad{\partial\bar z\over\partial y}=-i\ ,\tag{2}$$ whereas on the right hand side we obtain, using the CR equations, that $${\partial f\over\partial x}=u_x+iv_x,\qquad {\partial f\over\partial y}=u_y+iv_y=-v_x+iu_x\ .\tag{3}$$ If we assume $(1)$ then comparing the first equations of $(2)$ and $(3)$ leads to $u_x=1$, $v_x=0$, and comparing the second equations leads to $u_x=-1$, $v_x=0 \ $ – a contradiction.

Christian Blatter
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