# I Analytic functions of analytic functions

1. Jun 10, 2016

### Physgeek64

In our complex variables course we were told that an analytic function of an analytic function is itself analytic. i.e. For $h(z)=g(f(z))$ $h(z)$ is analytic.

I was wondering is this is just a fact, or if it is possible to prove this statement. I did some googling and the best response I could find was :

$h'(z)= g'(f(z)) f'(z)$, which for $f'(z)$ and $g'(f(z))$ not equal to zero, describes an analytic function.. But I'm afraid this doesn't seem like much of a proof to me.

Many thanks :)

2. Jun 10, 2016

### Staff: Mentor

What is your definition of an analytic function? Is it
"We say that the function $f$ is analytic in a neighbourhood $U$ of $z_0$ if it is differentiable everywhere in $U$." (http://www.maths.ed.ac.uk/~jmf/Teaching/MT3/ComplexAnalysis.pdf)
or is it
"In mathematics, an analytic function is a function that is locally given by a convergent power series."
(https://en.wikipedia.org/wiki/Analytic_function)

For the first one the differentiation you mentioned would be enough (plus that $g(U)$ is open).
Then it remains to show that the two definitions are equivalent. (See chapter 2.3.3 in the first source I mentioned.)

3. Jun 10, 2016

### mathman

4. Jun 10, 2016

### andrewkirk

I suggest attempting to prove it as follows.

Choose an arbitrary point $z$ and a point $z'$ sufficiently close to $z$ that the power series $\Big(S^{f,z}_n(z')\Big)_{n=1}^\infty$ for $f(z')$ at $z$ converges to $f(z')$ and the power series $\Big(S^{g,f(z)}_n(z')\Big)_{n=1}^\infty$ for $g(f(z'))$ at $f(z)$ converges to $g(f(z'))$. We know this can be done by the definition of analyticity of the two functions.

Then take the power series for $g(f(z'))$ and replace all occurrences of $f(z')$ by the power series for $f(z')$. If we write $\Delta z'\equiv z'-z$ then this gives us an infinite series whose terms are powers of an infinite series in $\Delta z'$. By adding coefficients across the same powers of $\Delta z'$ we can find the coefficient of each power of $\Delta z'$ in the power series that is our candidate for the actual power series for $h$. Write out the coefficients for the first few powers and a pattern should emerge that will allow writing a formula for the coefficient of $(\Delta z')^n$. We now have a power series that we would hope will converge to $h(z')$ for $z'$ sufficiently close to $z$.

Now we need to use epsilon-delta arguments to show that there is some $r$ such that for $|z'-z|<r$, the power series for $g(f(z'))$ does converge and is equal to $g(f(z'))$.

Both parts have their challenges, but it's an interesting yet eminently doable problem and should be enjoyable.