How are $z$ and $z^*$ independent?

Question

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 $y$.

Answer 1

It is true that the coefficients in $z$ and $z^*$ are related.

However, when we mean independence between $z$ and $z^*$, what we're actually saying is that $z$ and $z^*$ are $\mathbb{R}$-linearly independent (or linearly independent over $\mathbb{R}$), which means that for any $\alpha,\beta\in\mathbb{R}$ and $z\not\in\mathbb{R}$ and $z$ not on the imaginary axis, the only way to make $\alpha z + \beta z^*=0$ is to have $\alpha=\beta=0$.

Geometrically speaking, for a complex number $z$ that is not a real number, we do the following procedure within the complex plane: take $z$, scale it only by magnitude, then take $z^*$ and scale it by some other magnitude. The only way to make the sum of these two scaled complex numbers equal to zero, is to diminish both $z$ and $z^*$ to $0$ by scaling both $z$ and $z^*$ by $0$.

On the other hand, it should be stressed that the two complex numbers are NOT independent when we consider linear independence over the complex numbers $\mathbb{C}$. This is because $$z^*=\frac{z^*}{z} z,$$ and $z^*/z$ is a complex number that can scale $z$ to $z^*$.

---

Using terms of linear algebra, the difference results from the underlying field we choose.

Recall that each vector space is defined over a field. When viewing $\mathbb{C}$ as a vector field over the field $\mathbb{C}$, any basis only contains one element, the dimension of this vector field is 1. However, when viewing $\mathbb{C}$ as a vector field over the field $\mathbb{R}$, any basis contains two linearly independent elements, the dimension of this vector field becomes 2. When $z$ is not a real number and also not a real multiple of $i$, $z$ and $z^*$ serve as a basis for $\mathbb{C}_{\mathbb R}$, which is a short hand for $\mathbb{C}$ over $\mathbb{R}$.

Answer 2

At least in physics, the statement that $z=\alpha + \beta i$ and $z^*$ are independent relies on the generalization of $\alpha$ and $\beta$ to complex numbers, which unfortunately often goes without saying. However, one can almost always assume so and take $\alpha,\beta \in \mathbb{Z}$ at the end.

Answer 3

It is true that there is a one-to-one map between $z$ and $z^*$, it's just the reflection about the $x$-axis of the complex plane. Therefore, it is certainly not true that $z$ and $z^*$ are independent. However, if we consider $z^*$ as a function of $z$, so that $z^* = f(z)$, then it turns out that $f(z)$ is not a "nice" function in a sense that it cannot be built out of basic arithmetic operations such as $+,-,\times,\div$ and, finally, it is not differentiable. This means that, in the complex setting, the complex conjugation becomes an additional, independent, "arithmetic operation". It is in this sense that $z$ and $z^*$ are independent.

Answer 4

In several comments, OP mentions the quantity $\frac{dz^*}{dz}$, so I think it would be useful to clear something up.

If $f:\mathbb{R}\rightarrow \mathbb{R}$ is a function on the real numbers, then we can define its derivative (assuming it exists) as $$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$ Recall that for the limit to be well-defined, we must arrive at the same value whether $h\rightarrow 0$ from the left or from the right.

---

If $f:\mathbb{C}\rightarrow \mathbb{C}$ is a function on the complex numbers, then we similarily define its derivative as $$ f'(z) = \lim_{h \rightarrow 0} \frac{f(z+h)-f(z)}{h} $$

Notice that there is a subtle difference here. Now $h$ is complex, so there are an infinity of directions along which it could potentially go to zero. For this limit to be well-defined, we must arrive at the same value for all of these cases.

Applying this to the complex conjugate map, $$ \frac{dz^*}{dz} = \lim_{h\rightarrow 0}\frac{(z+h)^* - z^*}{h} = \lim_{h\rightarrow 0} \frac{h^*}{h}$$

The issue becomes clear when we write $h=|h|e^{i\theta}$ and $ h^*=|h|e^{-i\theta}$. The derivative becomes

$$\frac{dz^*}{dz} = e^{-2i\theta}$$

If $h$ is purely real ($\theta=0$), then $\frac{dz^*}{dz}=1$. If $h$ is purely imaginary ($\theta=\pi/2$), then $\frac{dz^*}{dz}=-1$. Obviously the value of the derivative of the complex conjugate map depends on the direction of the infinitesimal displacement $dz$. This is no good - it means that $\frac{dz^*}{dz}$ is not well-defined, and so $z^*$ is not a complex-differentiable function of $z$.

It follows from this that if a complex map is a function of both $z$ and $z^*$ (say, $f(z)= \mathbb{Re}(z) = \frac{1}{2}(z+z^*)$), then it is only complex-differentiable if you consider it a function of two independent variables and differentiate accordingly. In other words, we can say that $$ f = f(z,z^*)$$ $$\frac{\partial f}{\partial z} = \frac{1}{2}, \frac{\partial f}{\partial z^*} = \frac{1}{2}$$

but $\frac{df}{dz}$ is not a meaningful notion.

Answer 5

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.