Question on derivative of conjugate

[dimension10]

I was trying to find the derivative of $\overline{x}$ for some $x \in \mathbb{C}$

I solved this as

$$\frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h}-\overline{x}}{h}$$

$$\frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h-x}}{h}$$

$$\frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{h}}{h}$$

Now, am I right to say that that we can only simplify this further if we know $\mbox{arg}(\mbox{d}x)$, that is the angle of dx?

Thanks.

 

[micromass]

The function $f(x)=\overline{x}$ isn't differentiable anywhere.

 

[jackmell]

The function $f(x)=\overline{x}$ isn't differentiable anywhere.

I do not think it's correct to write:

$$f(x)=\overline{x}$$

rather I think it should be written

$$f(x,\overline{x})=\overline{x}$$

and then:

$$\frac{\partial f}{\partial x}=0$$

and therefore it's derivative with respect to x exists everywhere. However, since:

$$\frac{\partial f}{\partial \overline{x}}=1$$

it's not analytic anywhere.

 

[micromass]

I do not think it's correct to write:

$$f(x)=\overline{x}$$

rather I think it should be written

$$f(x,\overline{x})=\overline{x}$$

Huh, why? That makes no sense.

We are talking about the complex conjugation here...

 

[jackmell]

Huh, why? That makes no sense.

We are talking about the complex conjugation here...

Well, I didn't wanna' say this initially cus' you'll just hate me, but I feel writing

$$f(x)=\overline{x}$$

is like writing

$$f(a)=b$$

May I request a third opinion about the matter?

 

[micromass]

Well, I didn't wanna' say this initially cus' you'll just hate me,

Why would I hate you??

but I feel writing

$$f(x)=\overline{x}$$

is like writing

$$f(a)=b$$

Do you have anything against:

$$f:\mathbb{R}^2\rightarrow \mathbb{R}^2:(a,b)\rightarrow (a,-b)$$

And what about

$$f:\mathbb{C}\rightarrow \mathbb{C}:a+bi\rightarrow a-bi$$

and what about

$$f:\mathbb{C}\rightarrow \mathbb{C}:x\rightarrow \overline{x}$$

??

May I request a third opinion about the matter?

Sure, we'll see if anybody else responds.

 

[jackmell]

Those are good points. I'm not sure then. I think I'm right though and I'll stick by my guns and write what I stated above as the answer on my test paper.

Edit:

Now I think about it, I'm probably wrong because if the derivative were zero everywhere then that means it must be analytic. I don't understand it then. Sorry for getting involved.

I appologize Micromass.

 

[I like Serena]

Third opinion.The function:
$$f: \mathbb{C} \rightarrow \mathbb{C}$$
defined by:
$$f(z) = \bar z$$
is also given by:
$$f(a+bi) = a - bi$$

That is, what micro said.

 

[micromass]

I think you're confused by the notation $\frac{\partial}{\partial \overline{z}}$. This isn't the notation for partial derivative with respect to the $\overline{z}$ direction!

In fact, take a function $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, we define

$$\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)$$

So this is the very definition of that notation! The notation $\frac{\partial}{\partial \overline{z}}$ does not siginify a partial deriviative in the conventional sense but rather a special linear operator: a Wirtinger derivative. See http://en.wikipedia.org/wiki/Wirtinger_derivatives

 

[jackmell]

Ok, thanks guys.

Sorry Dimension10.

 

[dimension10]

So does that mean that the conjugate is not a function? But why? I mean, $\lim_{h \rightarrow 0} \frac{\overline{h}}{h}$ can be solved knowing $\arg h$. For example, if

$$\arg h = \frac{\pi}{4},$$

In other words, h goes equally in the real and imaginary directions. Then,

$$\Re (h)=\Im (h)$$
$$\overline{h}=\Re (h) -i \Im (h)$$
$$\overline{h}=\Re(h) \left( 1-i \right)$$
$$h=\Re (h)+i \Im (h)$$
$$h= \Re (h) \left( 1+i \right)$$
$$\frac{\overline{h}}{h}=\frac{\Re(h) \left( 1-i \right)}{\Re(h) \left( 1+i \right)}$$
$$\frac{\overline{h}}{h}=\frac{1-i}{1+i}$$
$$\lim_{h \rightarrow 0} \frac{\overline{h}}{h}=\frac{1-i}{1+i}$$
$$\frac{\mbox{d}y}{\mbox{d}x}=\frac{1-i}{1+i}=\frac{{\left(1-i\right)}^{2}}{(1+i)(1-i)}$$
$$\frac{\mbox{d}y}{\mbox{d}x}=\frac{1-i}{1+i}=\frac{-2i}{2}$$
$$\frac{\mbox{d}y}{\mbox{d}x}=\frac{1-i}{1+i}=-i$$

So when $\arg h = \frac{\pi}{4}$, then the derivative of the conjugate is $-i$.

So that is one example of when this is differentiable.

 

[I like Serena]

Since the limit would have a different value depending on $\arg h$, that means that the limit does not exist.

Btw, directional derivatives (which you appear to be referring to) do exist.

 

[dimension10]

Since the limit would have a different value depending on $\arg h$, that means that the limit does not exist.

Btw, directional derivatives (which you appear to be referring to) do exist.

Yes, I am referring to directional derivatives.

 

[I like Serena]

Actually, I can't recall having seen directional derivatives in complex analysis before, but the notation with u=1+i (in your example) would be one of the forms:
$$\nabla_u f(z), \qquad {\partial f(z) \over \partial u}, \qquad f'_u(z), \qquad D_u f(z), \qquad u \cdot \nabla f(z)$$

I took these from the wikipedia page on directional derivatives:
http://en.wikipedia.org/wiki/Directional_derivative

 

[dimension10]

Actually, I can't recall having seen directional derivatives in complex analysis before, but the notation with u=1+i (in your example) would be one of the forms:
$$\nabla_u f(z), \qquad {\partial f(z) \over \partial u}, \qquad f'_u(z), \qquad D_u f(z), \qquad u \cdot \nabla f(z)$$

I took these from the wikipedia page on directional derivatives:
http://en.wikipedia.org/wiki/Directional_derivative

Shoudn't

$$u=\frac{1+i}{\sqrt{2}}$$

Since, by definition, $\left| u \right|=1$ or in other words, $u={e}^{i\theta}$ for some $\theta\in\mathbb{R}$?

 

[I like Serena]

Depends.

From wikipedia: "One sometimes permits non-unit vectors, allowing the directional derivative to be taken in the direction of u, where u is any nonzero vector. In this case, one must modify the definitions to account for the fact that u may not be normalized."

 

[dimension10]

\begin{align}
& \arg h=\theta \\
& \theta \in \mathbb{R} \\
& \frac{\Im \left( h \right)}{\Re \left( h \right)}=\tan \theta \\
& h=\Re \left( h \right)+i\Im \left( h \right) \\
& \bar{h}=\Re \left( h \right)-i\Im \left( h \right) \\
& h=\Re \left( h \right)+i\Re \left( h \right)\tan \theta \\
& \bar{h}=\Re \left( h \right)-i\Re \left( h \right)\tan \theta \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{\bar{h}}}{h} \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\Re \left( h \right)-i\Re \left( h \right)\tan \theta }{\Re \left( h \right)+i\Re \left( h \right)\tan \theta } \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\Re \left( h \right)\left( 1-i\tan \theta \right)}{\Re \left( h \right)\left( 1+i\tan \theta \right)} \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-i\tan \theta }{1+i\tan \theta } \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( 1-i\tan \theta \right)}^{2}}}{\left( 1+i\tan \theta \right)\left( 1-i\tan \theta \right)} \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-2i\tan \theta -{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\frac{1-2i\tan \theta -{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta } \\
& \frac{\text{d}}{\text{d}x}\bar{x}={{\cos }^{2}}\theta \left( 1-2i\tan \theta -{{\tan }^{2}}\theta \right) \\
& \frac{\text{d}}{\text{d}x}\bar{x}={{\cos }^{2}}\theta -2i{{\cos }^{2}}\theta \tan \theta -{{\cos }^{2}}\theta {{\tan }^{2}}\theta \\
& \frac{\text{d}}{\text{d}x}\bar{x}={{\cos }^{2}}\theta -2i\sin \theta \cos \theta -{{\sin }^{2}}\theta \\
& \frac{\text{d}}{\text{d}x}\bar{x}=\cos 2\theta -i\sin 2\theta \\
& \frac{\text{d}}{\text{d}x}\bar{x}={{e}^{-2i\theta }} \\
\end{align}

 

[dimension10]

Depends.

From wikipedia: "One sometimes permits non-unit vectors, allowing the directional derivative to be taken in the direction of u, where u is any nonzero vector. In this case, one must modify the definitions to account for the fact that u may not be normalized."

Ok, but non-unit vectors? I just don't get how scalars like complex numbers can be vectors. Isn't it more of a vector that corresponds to the scalar rather than the scalar itself being a vector?

 

[I like Serena]

Since you write ${d \over dx} \bar x$, you are talking about the regular derivative and not the directional derivative.

And if you write $\lim\limits_{h \to 0} {\bar h \over h}$, obviously h is a complex number, so you are talking about the regular limit for complex numbers and not the directional limit.


Since you mean the directional derivative, you should write ${\partial \over \partial u} \bar x$.
And: $\lim\limits_{h \to 0} {h\bar u \over h}$, where h is a real number and u is a complex number with unit length.


Btw, I prefer $z$ instead of $x$, since $x$ suggests it's a real number, while $z$ usually represents a complex number.

 

[I like Serena]

Ok, but non-unit vectors? I just don't get how scalars like complex numbers can be vectors. Isn't it more of a vector that corresponds to the scalar rather than the scalar itself being a vector?

Yeah, the page is actually about vectors.
But complex numbers are a bit ambiguous.
Since they are elements of a field, they are scalars.
But often they are identified with vectors in $\mathbb{R}^2$ and if they are (which is what you are doing), they need to be treated as vectors.

Since we're really talking about directional derivatives, you should read "complex numbers" instead of "vectors" in that text.

 

[dimension10]

Since you write ${d \over dx} \bar x$, you are talking about the regular derivative and not the directional derivative.

And if you write $\lim\limits_{h \to 0} {\bar h \over h}$, obviously h is a complex number, so you are talking about the regular limit for complex numbers and not the directional limit.Since you mean the directional derivative, you should write ${\partial \over \partial u} \bar x$.
And: $\lim\limits_{h \to 0} {h\bar u \over h}$, where h is a real number and u is a complex number with unit length.Btw, I prefer $z$ instead of $x$, since $x$ suggests it's a real number, while $z$ usually represents a complex number.

Ok, but is it valid to define an operator like

\begin{align}
& \xrightarrow[D]{}=\frac{\partial }{\partial u} \\
& \text{for}\left| u \right|=1 \\
\end{align}

Here, putting an arrow does not denote a vector but just direction in the complex plane.

Also, shouldn't it be

$$\frac{\partial}{\partial (ux)}$$

instead since the bottom should be a variable? Or is it just a convention?

Also, is the following notation valid:

$$ \lim_{h\rightarrow {0}^{u}} \frac{\overline{h}}{h} $$
for
$$ |u|=1 $$

Asking because if $u=\pm 1$ then, I've seen this notation being used...

Thanks.

 

[dimension10]

But often they are identified with vectors in $\mathbb{R}^2$ and if they are (which is what you are doing), they need to be treated as vectors.

I don't see how I'm identifying them on $\mathbb{R}^{2}$? By direction, I do mean complex argument don't I? So they're still on $\mathbb{C}$. It just may be on $\mathbb{R}$ or $\mathbb I$.

For example, the dot product of 2 scalars is their product, right? So, since a complex number is a scalar,

$$ \begin{bmatrix}
x\\
y
\end{bmatrix}\in\mathbb{C}^2 $$

$$ x=a+ib $$
$$ y=p+iq $$
for some
$$ \begin{bmatrix}
a\\
b\\
p\\
q
\end{bmatrix}\in\mathbb{R}^{4} $$

$$ x\cdot y=xy=(a+ib)(p+iq)=ap-bq+i(aq+bp) $$

But

$$ \begin{bmatrix}
a\\
b
\end{bmatrix}\cdot\begin{bmatrix}
p\\
q
\end{bmatrix}=a\overline{p}+b\overline{q}=ap+bq \mbox{ since } \begin{bmatrix}
a\\
b\\
p\\
q
\end{bmatrix}\in\mathbb{R}^{4} $$

$ap-bq+i(aq+bp)=ap+bq$ only if $bq=0$ thus either $b$ or $q$ are $0$ which is not always true. Also, $aq+bp=0$ is another constraint which is also not necessarily true. So, complex numbers just can't be vectors.

 

[I like Serena]

Ok, but is it valid to define an operator like

\begin{align}
& \xrightarrow[D]{}=\frac{\partial }{\partial u} \\
& \text{for}\left| u \right|=1 \\
\end{align}

Here, putting an arrow does not denote a vector but just direction in the complex plane.

Sure, if you define it as a directional derivative with respect to u.
In math you can write anything you like, as long as you define it properly.
But why introduce yet another type of notation if there are already so many to choose from that people are familiar with?

Also, shouldn't it be

$$\frac{\partial}{\partial (ux)}$$

instead since the bottom should be a variable? Or is it just a convention?

It is a convention.
The regular partial derivative $\frac{\partial}{\partial x}$ can also be seen as the directional derivative with respect to $\mathbf{\hat x}$, the unit vector in the x direction.

Also, is the following notation valid:

$$ \lim_{h\rightarrow {0}^{u}} \frac{\overline{h}}{h} $$
for
$$ |u|=1 $$

Asking because if $u=\pm 1$ then, I've seen this notation being used...

Thanks.

Funny!
I had not realized this, but indeed 0+ and 0- also note directional derivatives for the real numbers.
I recognize the notation $\lim\limits_{h \to 0^+}$, which I actually know as $\lim\limits_{h \downarrow 0}$.
Here's a wiki page on one-sided limits for real numbers that shows even other possibilities: http://en.wikipedia.org/wiki/One-sided_limit

I guess you could do that too for complex numbers, although I wouldn't.
It's just not conventional.

 

[I like Serena]

I don't see how I'm identifying them on $\mathbb{R}^{2}$? By direction, I do mean complex argument don't I? So they're still on $\mathbb{C}$. It just may be on $\mathbb{R}$ or $\mathbb I$.

Again you are touching on the ambiguity of complex numbers where scalars and vectors are concerned.

A scalar is supposed to have no direction and be invariant under transformations.
But a vector does have direction which changes under transformations.

Part of the definition of a vector space is scalar multiplication, which is the multiplication of a scalar with a vector.
And we can additionally define scalar products of vectors that yield a scalar (inner products).
These scalars can be complex numbers, but they do not change the direction of the vector.

Depending how you treat a complex number and in which context you use it, a complex number can be either a scalar or a vector.
When you start talking about a direction, you are talking about a vector and not a scalar.

For example, the dot product of 2 scalars is their product, right? So, since a complex number is a scalar,

$$ \begin{bmatrix}
x\\
y
\end{bmatrix}\in\mathbb{C}^2 $$

$$ x=a+ib $$
$$ y=p+iq $$
for some
$$ \begin{bmatrix}
a\\
b\\
p\\
q
\end{bmatrix}\in\mathbb{R}^{4} $$

$$ x\cdot y=xy=(a+ib)(p+iq)=ap-bq+i(aq+bp) $$

But

$$ \begin{bmatrix}
a\\
b
\end{bmatrix}\cdot\begin{bmatrix}
p\\
q
\end{bmatrix}=a\overline{p}+b\overline{q}=ap+bq \mbox{ since } \begin{bmatrix}
a\\
b\\
p\\
q
\end{bmatrix}\in\mathbb{R}^{4} $$

$ap-bq+i(aq+bp)=ap+bq$ only if $bq=0$ thus either $b$ or $q$ are $0$ which is not always true. Also, $aq+bp=0$ is another constraint which is also not necessarily true. So, complex numbers just can't be vectors.

You are using the regular definition of a dot product here, which is indeed not the same for $\mathbb{C}^2$ and $\mathbb{R}^4$.
Those two vector spaces are isomorphic, but they have conventionally different dot products.

Note that the dot product (more generally the inner product) is not a part of the definition of a vector space, but is an additional structure on a vector space.

 

[dimension10]

Again you are touching on the ambiguity of complex numbers where scalars and vectors are concerned.

A scalar is supposed to have no direction and be invariant under transformations.
But a vector does have direction which changes under transformations.

Part of the definition of a vector space is scalar multiplication, which is the multiplication of a scalar with a vector.
And we can additionally define scalar products of vectors that yield a scalar (inner products).
These scalars can be complex numbers, but they do not change the direction of the vector.

Depending how you treat a complex number and in which context you use it, a complex number can be either a scalar or a vector.
When you start talking about a direction, you are talking about a vector and not a scalar.

You are using the regular definition of a dot product here, which is indeed not the same for $\mathbb{C}^2$ and $\mathbb{R}^4$.
Those two vector spaces are isomorphic, but they have conventionally different dot products.

Note that the dot product (more generally the inner product) is not a part of the definition of a vector space, but is an additional structure on a vector space.

If

$$ x\hat{\imath}+y\hat{\jmath}=x+iy $$

Then,

$$ \hat{\imath}=1 $$
$$ \hat{\jmath}=i $$
$$ \hat{\jmath} \hat{\jmath}=-1 $$
But if we use the geometric product of j hat and j hat, we get

$$ \hat{\jmath}\hat{\jmath}= \hat{\jmath} \wedge\hat{\jmath}+\hat{\jmath}\cdot\hat{\jmath}=0+1=1 $$

But, definitely, $-1\neq 1$...

 

[I like Serena]

Good point.

Regular multiplication of complex numbers is not a properly defined inner product when we treat complex numbers as vectors, since it violates positive-definiteness.

Now if we pick $\langle z_1, z_2 \rangle = z_1 \bar {z_2}$, I believe we have a proper inner product.

 

[Hurkyl]

I do not think it's correct to write:

$$f(x)=\overline{x}$$

rather I think it should be written

$$f(x,\overline{x})=\overline{x}$$

What you suggest is not a usual thing -- it is a trick that is sometimes used (a useful one, I suppose), taking a non-analytic function of x and writing it as an analytic function of $x$ and $\bar{x}$.

 

[Hurkyl]

But really, I think complex differential forms are more convenient when doing calculus with non-analytic functions. In one complex variable $z = x + iy$, you have two basis forms dx and dy. But it is often convenient to use a different basis, defined by
$$dz = dx + i dy$$
$$d\bar{z} = dx - i dy$$
And it can often be easy to write things out in this basis. e.g.

$$d|z| = d(z \bar{z})^{1/2} = \frac{1}{2} (z \bar{z})^{-1/2} d(z \bar{z}) = \frac{1}{2 |z|}(z d\bar{z} + \bar{z} dz)$$

 

[mathwonk]

the best R-linear approximation of a smooth map from C to C, is an element of the complex vector space HomR(C,C). (I.e. it is easy to multiply an R-linear transformation by a complex number, multiplying on the values.) The derivative in the real sense of a map from C to C, is such an R linear map (at each point).

This vector space has two natural subspaces, HomC(C,C), and HomCbar(C,C), the complex linear and conjugate linear maps, and they decompose HomR(C,C) into a direct sum. I.e. every real linear map C-->C can be written uniquely as a sum of a C-linear and a conjugate linear map.

In the case of a smooth map C-->C, the conjugate linear part of the real derivative, is called the ∂/∂zbar derivative, and the C-linear part is called the ∂/∂z derivative.

I.e. every R linear map from C to C can be written as a.z + b.zbar, with a, b complex constants. People who like derivatives to be numbers will call the coefficient a the ∂/∂z derivative, and the coefficient b, the ∂/∂zbar derivative.

We define a complex function to be holomorphic if its ∂/∂zbar derivative = 0. In the case of the function zbar, the ∂/∂z derivative = 0, and the ∂/∂zbar derivative equals 1.

To link this up with posts by Hurkyl and Micromass, notice that as a linear transformation, df = ∂f/∂x dx + ∂f/∂y dy.

Since dz+dzbar = 2dx, and dz-dzbar = 2idy, one can substitute for dx and dy in the expression for df and solve for the coefficients of dz and dzbar,

getting df = (1/2)(∂f/∂x -i ∂f/∂y) dz + (1/2)(∂f/∂x + i∂f/∂y) dzbar.

Thus ∂f/∂z, the coefficient of dz in the expression for df is what Micromass said. It is not however properly a definition, but a computation. Of course you can make it a definition, at the expense of trying to explain why you defined it that way.

This point of view makes some things obvious that are mentioned in that wiki article linked above, such as the laborious chain rule formulas there. I.e. it is trivial that in the composition of two R linear functions, the C-linear part of the composition arise from composing either both C-linear or both Cbar linear parts.

In fact the point of view given here is the original one, since it dates all the way back to Riemann, who expressed it on the second page of his inaugural dissertation in 1851, On the foundations of a general theory of functions of a complex variable. He writes out there the full expression for dw, in the form above, decomposing it into a dx+idy part, and a dx-idy part, i.e. into a dz and dzbar part.

He then distinguishes a class of functions he wishes to study, for which dw is always proportional only to dz, i.e. for which the coefficient of dx-idy is zero. This condition is of course the famous "Cauchy-Riemann" condition, that i∂f/∂x = ∂f/∂y.

 

[mathwonk]

In particular, in this, Riemann's original interpretation, and the interpretation now used in the theory of complex and almost complex manifolds, everything jackmell said originally is correct, up until he began to doubt himself.