Question on derivative of conjugate

1. Dec 28, 2011

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.

2. Dec 28, 2011

micromass

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

3. Dec 28, 2011

jackmell

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.

4. Dec 28, 2011

micromass

Huh, why???? That makes no sense.

We are talking about the complex conjugation here...

5. Dec 28, 2011

jackmell

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?

6. Dec 28, 2011

micromass

Why would I hate you??

Do you have anything against:

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

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

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

??

Sure, we'll see if anybody else responds.

Last edited by a moderator: Dec 28, 2011
7. Dec 28, 2011

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.

Last edited: Dec 28, 2011
8. Dec 28, 2011

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.

9. Dec 28, 2011

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

10. Dec 28, 2011

jackmell

Ok, thanks guys.

Sorry Dimension10.

11. Dec 28, 2011

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.

Last edited: Dec 28, 2011
12. Dec 28, 2011

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.

13. Dec 28, 2011

dimension10

Yes, I am referring to directional derivatives.

14. Dec 28, 2011

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:
$$\nabla_u f(z), \qquad {\partial f(z) \over \partial u}, \qquad f'_u(z), \qquad D_u f(z), \qquad u \cdot \nabla f(z)$$

http://en.wikipedia.org/wiki/Directional_derivative

15. Dec 28, 2011

dimension10

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 $\theta\in\mathbb{R}$?

16. Dec 28, 2011

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."

17. Dec 28, 2011

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}

18. Dec 28, 2011

dimension10

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?

19. Dec 28, 2011

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.

20. Dec 28, 2011

I like Serena

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.

21. Dec 29, 2011

dimension10

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.

Last edited: Dec 29, 2011
22. Dec 29, 2011

dimension10

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.

Last edited: Dec 29, 2011
23. Dec 29, 2011

I like Serena

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?

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.

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.

Last edited: Dec 29, 2011
24. Dec 29, 2011

I like Serena

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.

25. Dec 29, 2011

dimension10

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, $-1\neq 1$...