Infinitely differentiable complex functions

[ppy]

Hi
I am attempting to self-study Complex Analysis but i am confused over a couple of points.

1 - my book says "if a complex function is differentiable once throughout its domain of definition then it is infinitely differentiable" . How does this apply to z^2 ? If you differentiate it once you get 2z, differentiating again gives 2 and once more gives 0.

2- My book states that this infinitely differentiable point contrasts with the real case and gives an example of f(x)=0 for x≤0 and f(x) = x^2 for x≥0 and then goes on to say the second derivative at 0 does not exist, being 0 from the left and 2 from the right. But it seems to me that the same would happen for z^2 ?

Thanks

 

[Fredrik]

1 - my book says "if a complex function is differentiable once throughout its domain of definition then it is infinitely differentiable" . How does this apply to z^2 ? If you differentiate it once you get 2z, differentiating again gives 2 and once more gives 0.

And if you do it again, you get 0. Next time, 0. Next time, 0. Next time,...

You never get something like |z|, which is not differentiable.

2- My book states that this infinitely differentiable point contrasts with the real case and gives an example of f(x)=0 for x≤0 and f(x) = x^2 for x≥0 and then goes on to say the second derivative at 0 does not exist, being 0 from the left and 2 from the right. But it seems to me that the same would happen for z^2 ?

Hm, the f defined by
$$
f(z)=
\begin{cases}
0 & \text{if }\operatorname{Re} z<0\\
z^2 & \text{if }\operatorname{Re} z\geq 0
\end{cases}
$$ appears to be differentiable at 0, once, but not three times. So if we didn't both make the same mistake, you remember the theorem wrong. A quick look at Wikipedia suggests that this is the case.

A holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain. The existence of a complex derivative in a neighborhood is a very strong condition, for it implies that any holomorphic function is actually infinitely differentiable and equal to its own Taylor series.

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

The theorem doesn't ensure that our f is smooth, because it's differentiable only at the point 0, not at every point in some open set that contains 0. Try applying the definition of "differentiable" to this f at some non-zero point on the imaginary axis.

Note that in the real case, the function is differentiable on all of ℝ, not just at 0.

 

[ppy]

Thanks for your reply but I'm still confused on 2nd point.
If I differentiate z^2 once I get 2z. Differentiating again gives 2 so at the point 0 I have 2 different derivatives ; 0 and 2 so the derivative doesn't exist as in the real case ?

PS the book I'm using is "functions of a complex variable" by D.O. Tall

 

[Fredrik]

Thanks for your reply but I'm still confused on 2nd point.
If I differentiate z^2 once I get 2z. Differentiating again gives 2 so at the point 0 I have 2 different derivatives ; 0 and 2 so the derivative doesn't exist as in the real case ?

That's right. Both in the real case and in the complex case, the first and second order derivatives exist, but the third order derivative doesn't.

 

[ppy]

Thanks again but my book says that complex analytic functions are always infinitely differentiable in contrast to real functions but the above example shows that x^2 and z^2 are both not infinitely differentiable if we have f(x)=0 and f(z)=0 for x<0 and Re(z)<0

 

[Fredrik]

Thanks again but my book says that complex analytic functions are always infinitely differentiable in contrast to real functions but the above example shows that x^2 and z^2 are both not infinitely differentiable if we have f(x)=0 and f(z)=0 for x<0 and Re(z)<0

Do you mean "complex differentiable"? In that case, you are either remembering or interpreting the theorem wrong. See my comment in post #2, including the Wikipedia quote. The result we found doesn't contradict the theorem that says that functions that are complex differentiable on an open set are complex analytic on that set, because {0} is not an open set.

 

[Office_Shredder]

The function f(z) = z² for Re(z) > 0, 0 for Re(z) <=0 is not even continuous at, any point on the imaginary axis except for 0. The infinite differentiability theorem you're quoting relies on the domain of definition being open, and this is a good example of why (if you cut out just the imaginary axis but leave the origin, then your domain is not open)

 

[mathman]

There seems to be some confusion about the function under discussion.
f(z) = z² for ALL z is analytic.
f(x) = x² for x >0 and f(x) = 0 otherwise is not.