A Challenging Problem in Complex Analysis

Question

I want to prove or disprove by giving a counterexample: There is no entire function $f$ such that $f(R)\subset R$ and $f(i)=f(-i)=2019i$.

I appreciate any help in this problem as I have no clue about this problem.

Answer 1

Just adding some details to the comment by Count Iblis: for any entire fucntion $f$ the function $g$ defined by $g(z)=\overset {-} {f(\overset {-} {z})}$ is also an entire function (as seen from power series expansion). In this case $f=g$ or $f-g=0$ on $\mathbb R$. Since $\mathbb R$ has limit points in $\mathbb C$ it follows that $f-g=0$ everywhere. Now put $z=i$ in this to get a contradiction.