Is a complex function complex diffferentiable

1. Jun 20, 2009

lolgarithms

Is a complex function complex diffferentiable if AND only if they are analytic? or are there counterexamples?(analytic functions which are not holomorphic)

Why do you need to sometimes analyticially continue an analytic function?

Last edited: Jun 20, 2009
2. Jun 20, 2009

Civilized

Re: holomorphic=analytic?

That's right, this is one of the beautiful things which occurs in complex analysis.

In a higher dimensional complex space, the holomorphic functions are a proper subset of the analytic ones, but in the complex plane the two classes of functions coincide, this is a major theorem in complex analysis.

If the analytic function is defined not on the whole of the complex plane but only on an open subset.

3. Jun 20, 2009

zpconn

Re: holomorphic=analytic?

All analytic functions (real and complex) are infinitely differentiable. Thus any complex analytic function is complex differentiable, so it is holomorphic.

It's actually the converse that is interesting in complex analysis. Any function that is holomorphic is also analytic. Of course I'm speaking loosely here; consult a text on complex analysis to see precise statements of these results.

This is in stark contrast to the case in real analysis. For example, try to differentiate

f(x) = { 0 if x <= 0; x^2 if x >= 0 }

twice.

In fact, what's even more remarkable is that we can explicitly calculate all the derivatives of a holomorphic function. For example, if $$f : C \rightarrow C$$ is differentiable near $$x_0$$, then it is infinitely differentiable at $$x_0$$, and

$$f^{(n)}(x_0) = \frac{n!}{2\pi i} \int_{C} \frac{f(z)}{(z - x_0)^{n+1}}dz$$

where $$C$$ is just a sufficiently small loop surrounding $$x_0$$. Using this, one can explicitly compute the coefficients in the power series representation of a holomorphic function.

Last edited: Jun 20, 2009
4. Jun 20, 2009

Civilized

Re: holomorphic=analytic?

In the real numbers we can even have functions that are infinitely many times differentiable at a point that are still not analytic at that point e.g.

f(x) = {0 if x < 0 ; exp(-1/x^2) x > 0}

all the derivatives exist at x = 0, but they are all zero so the power series at that point is f(x) = 0 + 0 + 0 + ... which has zero radius of convergence and so the function is not analytic at that point.

5. Jun 20, 2009

lolgarithms

Re: holomorphic=analytic?

please give me an example of where analytic continuation is needed, and the power series for the analytically continued function.

Last edited: Jun 20, 2009
6. Jun 20, 2009

zpconn

Re: holomorphic=analytic?

Consider $$f(x) = 1/(1+x^2)$$ where x is real. If one writes

$$f(x) = \sum_{j=0}^{\infty} c_j X^j$$

where $$X = x - k$$ is the distance from x to some center k, then one can show that the radius of convergence is

$$\sqrt{1 + k^2}.$$

If we choose $$k = 0$$, for example, the radius of convergence is mysteriously 1, yet if one examines the graph of f one will not see anything wrong at x = 1 or x = -1.

In fact, this is when it is useful to consider the complex function $$F(z) = 1/(1+z^2)$$. Of course F(z) agrees with f(x) when z is restricted to the real line. Now note that the quantity

$$\sqrt{1+k^2}$$

is the distance in the complex plane from the center k to the points $$i$$ and $$-i$$. Of course, these points $$i, -i$$ are singularities of the complex function F(z) but not of the real function f(x), so we see that the radius of convergence of f(x) is the distance to the nearest singularity of the complex function F(z); in fact, although I won't explain it now, this is true generally if one replaces "nearest singularity" with "nearest singularity or branch point."

In this case I've "went backwards." Clearly F(z) is defined everywhere except at z = i and z = -i. So if one has a series representation of F(z) as above, then F(z) would be the analytic continuation of this series to the entire plane (except z = i and z = -i).

Here's a slightly better example. Consider the function

$$h(z) = 1 + z + z^2 + \dots$$

This converges within the unit disc $$|z| < 1$$ and consequently is analytic there. Except for a singularity at z = 1, this function can be analytically continued to the whole plane as

$$H(z) = \frac{1}{1-z}.$$

You can check that h and H agree within the unit disc. Further, this continuation is unique.

A very loose and rough way of thinking about this is as follows. If an analytic mapping sends any arbitrarily small curve to a point, then the mapping will send its entire domain to that point. Thus if $$H_1$$ and $$H_2$$ were two distinct continuations of $$h$$, then the function $$H_1 - H_2$$ would send the unit disc to the origin and thus $$H_1 - H_2 = 0$$.

7. Jun 25, 2009

g_edgar

Re: holomorphic=analytic?

On an open set, yes. At a single point, no.

Last edited: Jun 26, 2009