• Create Account

## Derivative of complex function: taken in which direction?

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

6 replies to this topic

### #1Lode  Members

1002
Like
0Likes
Like

Posted 27 April 2014 - 03:36 AM

Hi,

Given the function: f(x) = im(x). So, it takes the imaginary part of x.

On the real line, this function always returns 0. However, in the imaginary direction, it returns real values and has a slope of 1.

With the definition of the derivative, you look in the real direction only and miss that there's a slope in the imaginary direction. Wolfram Alpha gives 0: http://www.wolframalpha.com/input/?i=derivative+of+f%28x%29+%3D+im%28x%29+at+x+%3D+5

So, is that indeed correct then? Do you have to ignore the slope in imaginary direction for a derivative? Shouldn't the answer be something like 1, or -i?

My concern is that this derivative is not useful for Newton's method to find zeroes of complex functions, e.g. the above example would always return 0 even though there's a slope. I'm also wondering how to define slope in a 4D space (2D input, 2D output).

Thanks.

Edited by Lode, 27 April 2014 - 04:55 AM.

### #2Álvaro  Members

20265
Like
2Likes
Like

Posted 27 April 2014 - 05:14 AM

The complex functions that are "well behaved" for differentiations are called holomorphic. Basically, you need to have a limit of (f(z+h)-f(z))/h when h is a complex number that goes to 0. There are many ways to go to 0 in the complex plane, and all of them must give you the same answer. Newton's method works with holomorphic functions.

In your case, if you take a sequence of points that goes to 0 along the real axis you get a different answer than if you go to 0 along the imaginary axis, so you are not dealing with a holomorphic function, and Newton's method won't work.

In some sense, it's not useful to think of your function f as a complex function at all: It's more like a function from R^2 to R, because it ignores the complex structure in R^2.

### #3Lode  Members

1002
Like
0Likes
Like

Posted 27 April 2014 - 07:44 AM

Thanks! Very interesting. I did read a bit about holomorphic functions before, but failed to see it in this context.

One question still remains for me: Is the derivative of non-holomorphic functions defined at all, or is it simply considered "not differentiable"? And in particular, is the derivative of Im(x) simply the zero function?

I tried to search for it, but I find mostly things about holomorphic functions, not about non-holomorphic ones...

I'm experimenting with numerical math things, especially those that give pretty plots (like here http://www.gamedev.net/topic/655487-hsl-with-smoother-transitions/), and am trying to be generic so wouldn't want to exclude such functions.

Thanks!

### #4Álvaro  Members

20265
Like
1Likes
Like

Posted 27 April 2014 - 08:22 AM

If you think of your function as mapping R^2 to R, as I described earlier, you can define the derivative with respect to one of the two variables (this is called a partial derivative). So along the real direction the derivative is 0 and along the imaginary direction the derivative is 1. If you want to learn more about this, the general subject is multivariate calculus.

### #5Álvaro  Members

20265
Like
1Likes
Like

Posted 27 April 2014 - 11:15 AM

Oh, one more thing. You can still use Newton's method to find roots of a multivariate function, as explained here.

5832
Like
1Likes
Like

Posted 28 April 2014 - 04:45 AM

The complex derivative is not defined if the limit is different along different paths (also, you can't just consider straight line paths either, but I haven't got my analysis book to hand to give an example - the example in the book was a function that had a different limit along a parabola to the limit along a straight line path). So some functions which are differrentiable in R are not complex differentiable when embedded in C. If a function is complex differentiable at a point though, it is differentiable arbitrarily many times, which again is different to the case in R.

Wolfram Alpha gave the result for the partial derivative with respect to x (where z = x + iy). It didn't give a very satisfactory answer to derivative of f(z) = Im(z) either...

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

### #7Hotwire  Members

144
Like
1Likes
Like

Posted 07 May 2014 - 05:02 AM

There is such thing as Cauchy -Riemann constraint on complex function. If it satisfies it then you can you can find derivatives and as many as you wish.

It looks very easy if you go from

(x,y) to (z, \bar{z}).

In this basis constraint look as follows

d(f(z, \bar{z} )/d(\bar{z}) ==0

It simply means that your function doesn't depend on \bar{z}.