Finding $f(i)$ for an Entire Function $f$ that Maps a Line to a Subset of Itself and Sends $1$ to $0$. Consider the line $L=\lbrace x+iy: x=y\rbrace\subset \mathbb{C}$, suppose that $f:\mathbb{C}\to\mathbb{C}$ is entire and satisfies $f(L)\subset L$. Given $f(1)=0$, find $f(i).$ (Ans: $f(i)=0)$
I am not sure if I have covered the necessary material that will make this problem easier.
My ideas: 
1) Consider Taylor series' about points on the line or at $1,i$. Not sure why I thought of this as I cannot say much here.

2) Use the fact that $i$ is the reflection of $1$ about the line $y=x$. I don't recall learning anything that would allow me to take advantage of this fact, so it's just an observation.
How can I proceed? Ideally, I would like to be able to solve this problem for any line and for any two points, but I suppose the fact that the given points $1,i$ are symmetric with respect to the line is significant.
 A: How about $f(z) = z^2-1$? It sends the real line to a subset of the real line, (actually to a half-line); being a polynomial it  is entire.
EDIT, after seeing the OP's comment: Consider the family of polynomials $f_{n,a}(z) = a(z^n-1)$, with the parameter  $n$ chosen from multiples of 4 and $a$ any real number. All of them  take all 4th roots of unity to $0$, and the real line to  real line.
A: Any entire function can be written as
$$ g(z) =  \sum_{n=0}^\infty a_n z^n,\qquad a_n \in \mathbb{C}$$
A function which maps the reals to a subset of the reals will have coefficients $a_n \in \mathbb{R}$. Furthermore, we know that the composition of entire functions is entire. This suggests that if we can find entire functions $R, R'$ which map the line $\arg(z) = \pi/4$ to the real line, and the real line to the line $\arg(z) = \pi/4$, we can describe the functions which map the line $\arg(z) = \pi/4$ to a subset of itself.
The functions that we want are of course rotations in the complex plane. You can verify that if $\arg(z) = \pi/4$, then $e^{-i\pi/4} z $ will be real. Similarly, if $z$ is real, $e^{i\pi/4} z $ will lie on the line $y=x$.
This means that functions of the form $R' \circ g \circ R$ will map the line $y=x$ to a subset of itself. Here $g$ is any entire function with real coefficients in its power series.
Putting everything together, this means that we are looking at functions of the form
\begin{align*}
f(z) &= e^{i\pi/4}\sum_{n=0}^\infty a_n \left(ze^{-i\pi/4}\right)^n\\
&= e^{i\pi/4}\sum_{n=0}^\infty a_n z^n e^{-in\pi/4}
\end{align*}
with $a_n$ real.
Inserting that $f(1) = 0$, we see that
$$ 0 =   e^{i\pi/4}\sum_{n=0}^\infty a_n e^{-in\pi/4},$$
and writing $i = e^{i\pi/2}$, we have
\begin{align*}
f(i) &= e^{i\pi/4}\sum_{n=0}^\infty a_n e^{in\pi/2} e^{-in\pi/4}\\
&=e^{i\pi/4}\sum_{n=0}^\infty a_n e^{in\pi/4}\\
&= e^{i\pi/2}\left(e^{-i\pi/4}\sum_{n=0}^\infty a_n e^{in\pi/4}\right)\\
&= i \overline{f(1)}\\
&= 0
\end{align*}
A: Consider the function $g$ obtained from $f$ by turning by $45°$ in the target and domain (but opposite directions), that is set
$$g(z)=e^{-i\frac\pi4}f\left(e^{i\frac\pi4}z\right)$$
Then $g$ is entire and $g(\Bbb R)\subset \Bbb R$. The function $h(z)=\overline{g(\overline{z})}$ is entire aswell, and coincides with $g$ on $\Bbb R$, so they are equal on all of $\Bbb C$. Then 
$$0=e^{-i\frac\pi4}f\left(1\right)=g(e^{-i\frac\pi4})=h(e^{-i\frac\pi4})=\overline{g(e^{i\frac\pi4})}=e^{i\frac\pi4}f\left(i\right)$$
so that $f\left(i\right)=0$.
