Derivative of complex function Arg(z) or Re(z)+Im(z)?

MarieKaasi · New Member Offline Joined: 09 Feb 2005 Posts: 3

Please help me with this problem, I've tried everything!

How do I solve the derivative of functions f(z)=Arg(z) and g(z)=x+y (as z = x + iy) by using the definition. Meaning [f(z)-f(w)]/(z-w) -> f'(w) as z->w... And where is this derivative defined?

**Posted:** Wed Feb 09, 2005 11:22 am

JBL

1) This is a question of complex analysis. Very few high school students know complex analysis.

2) These look very much like homework problems -- do you have to do these for a class?

3) What do you mean, "Where is this derivative defined?" You gave the definition of the derivative within your post.

**Posted:** Wed Feb 09, 2005 11:59 am

_________________
Joel
Hi Deeps! <3

blahblahblah

Here are some hints.

If you know the Cauchy-Riemann equations, you have necessary and sufficient conditions for these derivatives to exist. Both of these functions are (continuous) functions from the complex numbers to the (strictly) real numbers.

In particular, the Cauchy-Riemann equations assert certain things about where functions from $\mathbb{C}$ to $\mathbb{R}$ can be differentiable. Besides that, I probably shouldn't say any more.

**Posted:** Wed Feb 09, 2005 12:05 pm

MarieKaasi · New Member Offline Joined: 09 Feb 2005 Posts: 3

>1) This is a question of complex analysis. Very few high school students know >complex analysis.

Here in Sweden this is high school-stuff.

>2) These look very much like homework problems -- do you have to do these for a >class?

These are not my homework. I'm studying for exam and these are basic exercises in my studybook - still I don´t know how to solve these. I think I should figure out how to simplify f(z)-f(w)/z-w so that z-w wouldn't approach to 0... just can´t do that!

>3) What do you mean, "Where is this derivative defined?" You gave the definition of >the derivative within your post.

I meant that for example the derivative of |x| is "defined" in R\{0}. Wrong word I guess.

And for what I've heard Arg(z) isn't continuous in negative x-axis, if it's defined (-pi,pi) as it normally is.[/b]

**Posted:** Wed Feb 09, 2005 1:58 pm

blahblahblah

Using the Cauchy-Riemann equations on $f(z)=\Re z+\Im z$ , we get

$u_x(x,y)=1$
$v_y(x,y)=0$

So right away the function is nowhere differentiable.

**Posted:** Wed Feb 09, 2005 2:22 pm

blahblahblah

**Posted:** Wed Feb 09, 2005 2:29 pm

koo · P versus NP Offline Joined: 27 Dec 2004 Posts: 24

I can infer that this is not HS level in Sweden.

**Posted:** Wed Feb 09, 2005 2:36 pm

DPatrick

Calculus of complex functions is not an "Intermediate Topic" for high-school students under any reasonable definition of "Intermediate".

Thus, this topic has been moved.

**Posted:** Wed Feb 09, 2005 3:10 pm

Kent Merryfield

If $f(z)=u(z)+iv(z)$ is smooth (i.e., has whatever properties of the existence and continuity of real derivatives you would want to ask for), then it is complex analytic if and only if is statisfies the Cauchy-Riemann equations:

$u_x=v_y$
$u_y=-v_x.$

If the function does satisfy these equations, then $f'=u_x+iv_x=v_y-iu_y.$

Now suppose $f$ takes on only real values. Note that this applies to both of your examples. Then $v$ =0 and thus $v_x=v_y=0$ everywhere. By the Cauchy-Riemann equations, we then have that $u_x=u_y=0,$ which makes $u$ and hence $f$ constant.

There are no non-constant real-valued complex analytic functions.

**Posted:** Wed Feb 09, 2005 3:49 pm

MarieKaasi · New Member Offline Joined: 09 Feb 2005 Posts: 3

Ok, now I got it!

So using the definition I can first approach by keeping Re(z) constant and then Im(z):

1) As Re(z)=Re(w): f(z)-f(w)/z-w = (Im(z)-Im(w))/i(Im(z)-Im(w)) = 1/i

But 2) As Im(z)=Im(w): f(z)-f(w)/z-w = (Re(z)-Re(w))/(Re(z)-Re(w)) = 1

=> f(z) isn't differentiable anywhere. Or is it in (0,0)?

Can this kind of approach be done somehow to the other function Arg(z)?

**Posted:** Thu Feb 10, 2005 5:10 am

Kent Merryfield

**Posted:** Thu Feb 10, 2005 8:17 am