Confusion about a strange implication of the Taylor series for complex numbers

Question

For all holomorphic $f$, $$f(z)=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}z^r.$$ Writing $z$ in polar form, we get $$f(z)=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}|z|^r(\cos(r\cdot\arg(z))+i\sin(r\cdot\arg(z))),$$ so $$\Re(f(z))=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}|z|^r\cos(r\cdot\arg(z)),$$ $$\Im(f(z))=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}|z|^r\sin(r\cdot\arg(z)).$$ Therefore, $\Re(f(z))$ is even with respect to $\arg(z)$ and $\Im(f(z))$ is odd with respect to $\arg(z)$, so $\Re(f(z_1))=\Re(f(z_2))$ and $\Im(f(z_1))=-\Im(f(z_2))$ when $z_1$ and $z_2$ are complex conjugates, for all holomorphic $f$. Is this actually true?? Seems very peculiar. The image of complex conjugates under a holomorphic function are always complex conjugates... It implies that the roots of any holomorphic function occur in conjugate pairs, for one thing. Have I gone wrong anywhere?

Answer 1

You are assuming that the coefficients $\frac {f^{r}(0)} {r!}$ are real. If $f(z)=\sum a_nz^{n}$ with $a_n$ real for all $n$ then it is true that $\Re(f(z))=\Re(f(\overset {-} z))$ (and similarly for the imaginary part). For $f(z)=iz$ these formulas are not valid.