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

[MarieKaasi]

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?

[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.

[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 to can be differentiable. Besides that, I probably shouldn't say any more.

[MarieKaasi]

>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]

[blahblahblah]

Using the Cauchy-Riemann equations on , we get




So right away the function is nowhere differentiable.

[blahblahblah]

[koo]

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

[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.

[Kent Merryfield]

If 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:




If the function does satisfy these equations, then

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

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

[MarieKaasi]

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)?

[Kent Merryfield]