# $f(z) = \sum_{n=0}^{\infty} c_n z^n$ has radius of convergence R. Show $\sum_{n=0}^{\infty} \overline{c_n} z^n$ has same radius of convergence

suppose $f(z) = \sum_{n=0}^{\infty} c_n z^n$ is a power series with finite radius of convergence $R > 0$. Show $\sum_{n=0}^{\infty} \overline{c_n} z^n$ has the same radius of convergence, R

I'm super stuck on this and I haven't learnt Hadamard's Theorem so please don't suggest it!

• Note that the second series is $\overline{f(\overline{z})}$. If $z$ is inside the convergence radius then so is $\overline{z}$. Mar 25 '18 at 22:52
• The radius of convergence is defined in terms of the modulus of the coefficients and $c_n,\overline{c_n}$ have the same modulus. There is no need to invoke Schwarz' reflection principle or Hadamard's theorem: your question is trivial by the very definition of radius of convergence. Mar 26 '18 at 0:00

The series $\sum_{n=0}^\infty c_nz^n$ converges if and only if the series $\sum_{n=0}^\infty\overline{c_n}\,\overline{z}^n$ converges. Besides, $\bigl|\overline z\bigr|=|z|$.