Laurent series of real part of holomorphic function

Question

In the book "Harmonic function theory" p. 9 there is this claim: let $f(z)=\sum_{j=0}^{+\infty}a_jz^j$ be a holomorphic function. Then its real part takes the form \begin{equation} u(r\xi)=\sum_{j=-\infty}^{+\infty}a_jr^{|j|}\xi^j,\qquad for \ 0<r<1, |\xi|=1. \end{equation} It does not convince me, here is my attempt.
Write $u(z)=\dfrac{f+\overline{f}}{2}(z)$ and $\xi=e^{it}$, then $$ f(re^{it})=\sum_{j=0}^{+\infty}a_jr^je^{ijt}=\sum_{j=0}^{+\infty}a_{|j|}r^{|j|}e^{ijt}=\sum_{j=0}^{+\infty}a_{|j|}r^{|j|}\xi^j $$ and $$ \overline{f}(re^{it})=\sum_{j=0}^{+\infty}\overline{a}_jr^je^{-ijt}=\sum_{k=-\infty}^{0}\overline{a}_{|k|}r^{|k|}e^{ikt}=\sum_{k=-\infty}^{0}\overline{a}_{|k|}r^{|k|}\xi^k, $$ but the equality does not follow.

Answer 1

They never wrote $f(z)=\sum_{j=0}^{+\infty}a_jz^j.$

Instead, let$$f(z)=\sum_{j=0}^{+\infty}b_jz^j.$$ Then, $$u(r\zeta)=\sum_{j=-\infty}^{+\infty}a_jr^{|j|}\zeta^j\quad \text{ for }\quad 0\le r<1, |\zeta|=1$$ where $$a_j=\begin{cases}\frac{b_j}2&\text{if }j>0\\\frac{\overline{b_{-j}}}2&\text{if }j<0\\\frac{b_0+\overline{b_0}}2&\text{if }j=0. \end{cases}$$