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Let $D=D(0,1)$ be the unit disc centered at 0. Let $f$ be holomorphic in D, with $f(0)=0$. Show that

$g(z)=\sum_{k=1}^{\infty} f(z^k)$

defines an holomorphic function in $D$.

I've argued as follows: since $f$ is holomorphic, then $f$ is locally the sum of a power series

$f(z)=\sum_{n=0}^{\infty}a_n z^n$

and since $f(0)=a_0=0$ by assumption, we can write

$f(z)=\sum_{n=1}^{\infty}a_nz^n$

which holds for all $z$ in $D$.

Hence

$g(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_n z^{nk}$

How can i deal with this double index summation?

$g(z)=\sum_{k=1}^{\infty} f(z^k)$

defines an holomorphic function in $D$.

I've argued as follows: since $f$ is holomorphic, then $f$ is locally the sum of a power series

$f(z)=\sum_{n=0}^{\infty}a_n z^n$

and since $f(0)=a_0=0$ by assumption, we can write

$f(z)=\sum_{n=1}^{\infty}a_nz^n$

which holds for all $z$ in $D$.

Hence

$g(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_n z^{nk}$

How can i deal with this double index summation?

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