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 $y$.

  • 3
    $\begingroup$ Perhaps who told you this, meant linearly independent (over $\Bbb R$). Nevertheless, if it is the case, it is true only if $z\notin\Bbb R$. $\endgroup$
    – ajotatxe
    Oct 4, 2017 at 21:24
  • $\begingroup$ Although $z$ and $z^{\ast}$ are not independent, it seems relevant to mention that in many applications, one can get away with treating them as independent, cf. this Phys.SE post $\endgroup$
    – Qmechanic
    Oct 4, 2017 at 21:45
  • $\begingroup$ This is EXACTLY the reason I am posting this question, because of this problem in my quantum field theory class! $\endgroup$ Oct 4, 2017 at 22:18

5 Answers 5


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 $\mathbb{R}$.

  • $\begingroup$ Sorry for deleting the comment: for posterity, it was: "Am I correct in saying that if $x$ and $y$ are real, then we have linear independence over $\mathbb R$, but not total independence? By total independence, I mean "if I know only one, I cannot know the other". On the other hand, if $x$ and $y$ are complex, then $z$ and $z*$ are totally independent of each other. And in neither case are $z$ and $z*$ linearly independent over $\mathbb C$." $\endgroup$ Oct 4, 2017 at 21:50
  • $\begingroup$ @Danny You're right in "Am I correct... over $\mathbb{R}$" and "And in neither case... over $\mathbb{C}$. By "total dependence", I think what you mean is that we can define a function (or also called a map, mapping) such that you can determine the result from the input. In this case, the $x$ and $y$ you chose are random from the beginning, so there is no real way of knowing one from the other (they are independent variables). In the case of complex conjugation, you can uniquely define $z^*$ from $z$. Any two complex numbers are dependent in $C$ even if they are not conjugate of each other. $\endgroup$
    – Frenzy Li
    Oct 4, 2017 at 21:51
  • $\begingroup$ @Danny See Fizikus' answer which offers another view on your "total dependence" by formulating it in terms of isomorphisms (one-to-one and onto maps). $\endgroup$
    – Frenzy Li
    Oct 4, 2017 at 21:54
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    $\begingroup$ Isn’t this claim false if z is pure imaginary? $\endgroup$ Oct 4, 2017 at 22:04
  • $\begingroup$ This is starting to make sense, but it is opening up even more contradictions in my head. If there is a mapping $f: \mathbb C \rightarrow \mathbb C$ defined by $f(z) = z*$, then $\frac{\partial z*}{\partial z}$ should either not exist or exist and not be identically zero. (For if it were identically zero, then $z*$ does not depend on $z$ which goes against are assumption that there is a mapping.) But for $z = x + iy$ and $z*=x-iy$, we can evaluate this partial derivative using the chain rule and find it is 0. I will read Fizikus' response. $\endgroup$ Oct 4, 2017 at 22:07

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.

  • $\begingroup$ Yes. This is the formalism we are usually taught in physics. $\endgroup$ Oct 5, 2017 at 22:37

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.

  • $\begingroup$ If it is not differentiable, then what is wrong with this? Let $z=x+iy$ for real $x$ and $y$. Then $\frac{\partial z*}{\partial z} = \frac{\partial z*}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial z*}{\partial y}\frac{\partial y}{\partial z} = 1\times \frac{1}{2} + (-i)\times\frac{1}{2i} = 0$. In physics, I have seen this used as an explanation for the independence of $z$ and $z*$. $\endgroup$ Oct 4, 2017 at 22:13
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    $\begingroup$ The notion of differentiability is much stronger than merely possesing partial derivatives. Differentiability is defined as in J. Murray's answer. $\endgroup$
    – Fizikus
    Oct 5, 2017 at 6:29

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.

  • $\begingroup$ Maybe this is an artifact of the typesetting, but I did not refer to $\frac{df}{dz}$; I only ever used partials, which imply that the variations over $z$ and $z*$ are independent. $\endgroup$ Oct 5, 2017 at 4:37
  • $\begingroup$ @Danny The map $f:\mathbb{C}\rightarrow \mathbb{C}$ such that $f(z)=z^*$ is a function of a single complex variable, so the partial derivative doesn't make much sense to use; if you do choose to use it, then you would have $\frac{\partial f}{\partial z} = \frac{df}{dz}$, and this object is undefined. $\endgroup$
    – J. Murray
    Oct 5, 2017 at 4:54
  • $\begingroup$ On the other hand, the map $g : \mathbb{C}\times\mathbb{C} \rightarrow \mathbb{C}$ such that $g(z,z^*)=z^*$ is a function of two complex variables. In this case, $\frac{\partial g}{\partial z}=0$ and is perfectly well-defined. $\endgroup$
    – J. Murray
    Oct 5, 2017 at 4:55
  • $\begingroup$ See Mike Stone's response below. I guess I am referring to Wirtinger derivatives. $\endgroup$ Oct 5, 2017 at 22:39

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.


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