# Continuation of a real function into the total complex plane

1. Mar 19, 2010

### wdlang

suppose i have a real function f=f(x)

this function is smooth everywhere on the real line

for example, f=e^x.

The problem is, is the continuation of the function into the complex plane unique?

if so, does it hold that f(z)=f(z*)*?

2. Mar 19, 2010

### wdlang

i just note that for function f=1/(1+x^2)

there must be a pole at z=i and -i

3. Mar 19, 2010

### dalle

dear wdlang, your function needs to be real analytic, being smooth is not enough. There is an excelent discussion of the difference by Dave Renfro at http://mathforum.org/kb/thread.jspa?forumID=13&threadID=81514&messageID=387148#387148

if the radius of convergence is infinite, than your function is entire, and the continuation is unique by the principle of permanence. As you note in your second post, things get complicated when the radius of convergence is finite.

let $$f_n = z^n$$. Is $$f_n(z)=f_n(z*)*$$?
let $$p$$ be an real polynomial. Is $$p(z)=p(z*)*$$
let $$g=\sum_{n=0}^{\infty }a_n z^n , a_n \in R$$ be a real power series with radius of convergence $$r > 0$$
let $$z \in C , |z| < r$$. Is $$g(z)=g(z*)*$$
Hint: Use that * is an continous automorphism of C.

4. Mar 19, 2010

### wdlang

yes, the function by cauchy is just amazing

exp(-1 / x^2) [with f(0) = 0]

the taylor series at x=0 does not converge to the function itself

5. Mar 19, 2010

### g_edgar

For uniqueness: yes... any two extensions agree on the real line, which is a set with a limit point, so they agree on their entire (connected) common domain.

f(z) = f(z*)* is the "reflection principle".

6. Mar 19, 2010

### dalle

for any other point x the radius of convergence is |x|. exp(-1/x^2) is holomorphic on C\{0} and the radius of convergence is the distance to the next singularity (which the function has only one at the point z=0)