Is a complex function complex diffferentiable

In summary: There are counterexamples.Please give me an example of where analytic continuation is needed, and the power series for the analytically continued function.In summary, if an analytic function is holomorphic, it is also analytic. If a function is analytic but not holomorphic, then it can be analytically continued but there may be a singularity at the point of continuation.
  • #1
lolgarithms
120
0
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?
 
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  • #2


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

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.

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

If the analytic function is defined not on the whole of the complex plane but only on an open subset.
 
  • #3


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 [tex]f : C \rightarrow C[/tex] is differentiable near [tex]x_0[/tex], then it is infinitely differentiable at [tex]x_0[/tex], and

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

where [tex]C[/tex] is just a sufficiently small loop surrounding [tex]x_0[/tex]. Using this, one can explicitly compute the coefficients in the power series representation of a holomorphic function.
 
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  • #4


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


please give me an example of where analytic continuation is needed, and the power series for the analytically continued function.
 
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  • #6


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

[tex]
f(x) = \sum_{j=0}^{\infty} c_j X^j
[/tex]

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

[tex]
\sqrt{1 + k^2}.
[/tex]

If we choose [tex]k = 0[/tex], 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 [tex]F(z) = 1/(1+z^2)[/tex]. Of course F(z) agrees with f(x) when z is restricted to the real line. Now note that the quantity

[tex]
\sqrt{1+k^2}
[/tex]

is the distance in the complex plane from the center k to the points [tex]i[/tex] and [tex]-i[/tex]. Of course, these points [tex]i, -i[/tex] 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

[tex]
h(z) = 1 + z + z^2 + \dots
[/tex]

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

[tex]
H(z) = \frac{1}{1-z}.
[/tex]

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 [tex]H_1[/tex] and [tex]H_2[/tex] were two distinct continuations of [tex]h[/tex], then the function [tex]H_1 - H_2[/tex] would send the unit disc to the origin and thus [tex]H_1 - H_2 = 0[/tex].
 
  • #7


lolgarithms said:
Is a complex function complex differentiable if AND only if they are analytic?

On an open set, yes. At a single point, no.
 
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1. What is a complex function?

A complex function is a mathematical function that operates on complex numbers, which are numbers that have both a real and imaginary part. Complex functions are typically represented by the form f(z), where z is a complex number.

2. How is complex differentiability defined?

A complex function f(z) is said to be complex differentiable at a point z0 if the limit of (f(z)-f(z0))/(z-z0) exists as z approaches z0. This limit is known as the complex derivative of f(z) at z0 and is denoted by f'(z0).

3. What is the difference between complex differentiability and real differentiability?

In real differentiability, the limit of (f(x)-f(x0))/(x-x0) exists as x approaches x0. However, in complex differentiability, the limit of (f(z)-f(z0))/(z-z0) must exist for all paths approaching z0, not just along the real axis.

4. How can I determine if a complex function is complex differentiable?

To determine if a complex function is complex differentiable, you can use the Cauchy-Riemann equations. If a function satisfies these equations, it is complex differentiable. However, if the equations are not satisfied, it does not necessarily mean the function is not complex differentiable.

5. Why is complex differentiability important?

Complex differentiability is important because it allows us to extend the concept of differentiation to complex functions, which are used in many areas of mathematics, physics, and engineering. This concept is also crucial in understanding and solving problems in complex analysis, a branch of mathematics that studies functions of a complex variable.

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