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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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.