TITLE: How are $z$ and $z^*$ independent?
QUESTION [7 upvotes]: I have been told that a complex number $z$ and its conjugate $z^*$ are independent. Part of me understands this, since for two independent variables $x$ and $y$ we can always define new independent variables $x' = \alpha x + \beta y$ and $y' = \alpha x - \beta y$.
However, this contradiction is confusing me:
Suppose I assume $x$ and $y$ are real. Then if I know $z$, I know both $x$ and $y$, which sort of makes sense because $\mathbb C \cong \mathbb R^2$. For example, if you tell me $z = 4 + 5i$, then $z^*$ is uniquely determined to be $4 - 5i$. How can we then say $z$ and $z^*$ are independent? I cannot change $z$ without also changing $z^*$. I can, however, change $x$ without changing $y$.

REPLY [5 votes]: I think the question raised by the OP is related to the Wirtinger calculus in which we write $z=x+iy$, $z^*=x-iy$ as usual, and then define
$$
\frac{\partial}{\partial z}\equiv  \frac 12 \left( \frac{\partial}{\partial x}- i \frac{\partial}{\partial y}\right)
$$
$$
\frac{\partial}{\partial z^*}\equiv  \frac 12 \left( \frac{\partial}{\partial x}+ i \frac{\partial}{\partial y}\right).
$$
The deinition is made so that with $dz=dx+idy$ and   $dz^*=dx-idy$ the variation 
$$
df= \frac{\partial f}{\partial x}dx
+ \frac{\partial f}{\partial y}dy
$$
becomes 
$$
df(z,z^*) = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial z^*}dz^*.
$$
From this we find 
$$
\frac{\partial}{\partial z^*}(z z^*) = z
$$
and so on, and hence that, for all practical calulational purposes, $z$ and $z^*$ can be treated as independent variables.