Let $f:\mathbb{C}\mapsto\mathbb{C}$ be holomorphic, then $$f(z)=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}z^r.$$ Writing $z\equiv|z|(\cos(\arg(z))+i\sin(\arg(z)))$, and using De Moivre's theorem, we get $$f(z)=\sum\limits_{r=0}^{\infty}\frac{f^{(r)}(0)|z|^r}{r!}(\cos(r\cdot\arg(z))+i\sin(r\cdot\arg(z))).$$ Therefore, $$\Re(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}(\Re(f^{(r)}(0))\cos(r\cdot\arg(z))-\Im(f^{(r)}(0))\sin(r\cdot\arg(z))),$$ $$\Im(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}(\Im(f^{(r)}(0))\cos(r\cdot\arg(z))+\Re(f^{(r)}(0))\sin(r\cdot\arg(z))).$$Suppose that $\Im(f^{(r)}(0))=0$ for all $r$, then $\Re(f(z))$ is even in $\arg(z)$, while $\Im(f(z))$ is odd in $\arg(z)$. Therefore, $f(\overline{z})=\overline{f(z)}$ for all $z\in\mathbb{C}$. Now suppose that $\Re(f^{(r)}(0))=0$ for all $r$, then $\Re(f(z))$ is odd in $\arg(z)$, while $\Im(f(z))$ is even in $\arg(z)$.
Let $f$ be such that $\Im(f^{(r)}(0))=0$ for all $r$, then whenever $f(z)=0$, $f(\overline{z})=0$. Now let $g(z)=i\cdot f(z)$ for all $z\in\mathbb{C}$, so $\Re(g^{(r)}(0))=0$ for all $r$. Since $\Re(g(z))$ is odd in $\arg(z)$ and $\Im(g(z))$ is even in $\arg(z)$, whenever $g(z)=0$, $g(\overline{z})=0$. But, $g$ is just the rotation of $f$ $\pi/2$ radians anti-clockwise about $0$, so whenever $g(z)=0$, $g(\hat{z})=0$, where $\hat{z}$ is the reflection of $z$ in the imaginary axis. But $\hat{z}=-\overline{z}$, so, if any holomorphic $f$ is such that $\Re(f^{(r)}(0))=0$ for all $r$, or $\Im(f^{(r)}(0))=0$ for all $r$, $f(-z)=0$ whenever $f(z)=0$. This is absurd. Where have I gone wrong?
Answer: After reading a comment I realised that $g(z)=0$ does not imply that $g(\hat{z})=0$ since it is the image of each point under $f$ that rotates when instead applying $g$ - my misunderstanding was that the zeroes of $g$ are located at the rotation of the zeroes of $f$, which is not true; the zeroes of $g$ are at the zeroes of $f$. In my head, was actually defining $g(iz)=f(z)$ without realising.