I have the following proof that a holomorphic function's zeroes occur in conjugate pairs when its derivatives evaluated at $0$ lie on a line through $0$:
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 $z\equiv\cos(\arg(z))+i\sin(\arg(z))$, and invoking De Moivre's theorem, we get $$\Re(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}(\Re(f^{(r)}(0))\cos(r\cdot\arg(z))-\Im(f^{(r)}(0))\sin(r\cdot\arg(z))),$$ $$\Im(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}(\Im(f^{(r)}(0))\cos(r\cdot\arg(z))+\Re(f^{(r)}(0))\sin(r\cdot\arg(z))).$$ Suppose $f^{(r)}(0)\in\mathbb{R}$ for all $r\in\mathbb{N}_0$, then $\Im(f^{(r)}(0))=0$ for all $r$, so $$\Re(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}f^{(r)}(0)\cos(r\cdot\arg(z)),$$ $$\Im(f(z))=\sum\limits_{r=0}^{\infty}\frac{|z|^r}{r!}f^{(r)}(0)\sin(r\cdot\arg(z)).$$ Therefore, $\Re(f(z))$ is even with respect to $\arg(z)$, and $\Im(f(z))$ is odd with respect to $\arg(z)$, so $f(\overline{z})=\overline{f(z)}$ for all $z\in\mathbb{C}$. Therefore, if $f(z)=0$, $f(\overline{z})=0$.
Now let $f$ be such that its derivatives evaluated at $0$ lie on a line through $0$ such that $\arg(z)=\theta$ for all non-zero $f^{(r)}(0)$, then $g(z)=e^{-i\theta}f(z)$ is such that $g^{(r)}(0)\in\mathbb{R}$ for all $r$ since $g^{(r)}(0)=e^{-i\theta}f^{(r)}(0)$. Therefore, the zeroes of $g$ occur in conjugate pairs, and $f$ is simply a rotation of $g$ about $0$, so the zeroes of $f$ occur in conjugate pairs.
My question is, are there known results about curves $\gamma$ in $\mathbb{C}$ other than lines through $0$ such that any holomorphic $f$ such that $f^{(r)}(0)\in\gamma$ for all $r\in\mathbb{N}_0$ is such that, if $f(z)=0$, $f(\overline{z})=0$. I've looked around online and all I can find is the conjugate pair theorem, which is a special case of what I have already proven.
A method I have tried of extending the class of curves that have this property is to note that, given some curve $\gamma$ and any sequence of points $(a_n)_{n\in\mathbb{N}}$ such that $a_n\in\gamma$ for all $n\in\mathbb{N}$, if there exists some holomorphic $g$ such that $g^{(r)}(0)\in\mathbb{R}$ for all $r$ and some holomorphic $T$ such that $T(0)=0$ and $T(z)\neq 0$ for all non-zero $z\in\mathbb{C}$, such that $(T\circ g)^{(r)}(0)=a_r$ for all $r$, then any function whose derivatives evaluated at $0$ lie on $\gamma$ is such that it zeroes occur in conjugate pairs. I won't write out the whole proof, but I then used the combinatorial form of Faa De Bruno's formula to show that one has sufficient freedom in the choosing of the derivatives of $T$ and $g$ to conclude that there exists some $T$ and $g$ such that $(T\circ g)^{(r)}(0)=a_r$ for all $r\in\mathbb{N}$, and, moreover, that there is even more freedom in the choosing than that; I have come close to proving that you can get $T(0)=0$, but not that $T(z)\neq 0$ for all non-zero $z$ (in fact, this is definitely not true for all $\gamma$, and I don't see how I will be able to distinguish between those curves for which it is true and those for which it isn't, even if I do make progress on determining what curves one has "more freedom" for), and was wondering if there are known results related to this.
