The function $\overline{z}$ has no power series expansion on a disk

Question

I'm confused as to how to prove this theorem, would appreciate any hints.

Answer 1

If $f$ has a power series expansion in some disk centered at the origin then $f$ would be holomorphic in that disk. As $f'(0)$ does not exist this can't happen.

Answer 2

If $f(z) = \overline{z}$ had a power series expansion at $0$ with positive radius of convergence $R$, then in particular that power series converges to $z$ when $z \in (-R,R)$ is real. The real function $g(x) = x$ has only one power series at $0$, namely $x$.$^{1}$ So the original power series for $f(z)$ must have been $z$, but then $z$ is not equal to $\overline{z}$ when $z$ is not real, contradiction.

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$^{1}$ This we know from real analysis. Specifically, we can show that a function can have at most one power series converging to it on an interval around $0$: if we suppose that the two power series $\sum_{i \ge 0} a_i x^i$ and $\sum_{i \ge 0} b_i x^i$ converge to the same function on an interval around $0$, then repeatedly taking derivatives we get $a_i = b_i$ for all $i$.

In particular $x$ can have at most one power series converging to it; since $x$ itself is a power series ($0 + x + 0x^2 + 0x^3 + \cdots)$, that is the only one.