Does there exist an entire function such that it transforms the real axis in the imaginary axis, and the imaginary axis into the parabola $y^2=x$?
I have been trying to solve this question for my complex analysis class, it's from another year exam.
I attempted to use the Cauchy-Riemman equations to prove that there is none, but having $f(x+iy)=u(x,y)+iv(x,y)$ resuts in having either $x=0$ or $y=0$ in my inputs of the funtion.
I end up having that:
(1) $u(x,0)=0$ and $v(x,0)$ is a surjective function;
(2) $u(0,y)^2=v(0,y)$;
(3) $u(0,0)=v(0,0)=0$ as it is the intersection of the images.
Then I attempted to prove that either $u$ or $v$ are not harmonic functions, but that led me nowhere for the same reason.
After trying multiple things I reach nowhere.
NOTE: An entire function is a holomorfic funtion in all the points of the complex plane.
If $f$ takes the real axis into the imaginary axis, then by the Schwarz reflection principle, $f(\overline{z})$ is the reflection of $f(z)$ across the imaginary axis, which is $- i \overline{i f(z)}$. If $f(z) = \sum_{j=0}^\infty c_j z^j$, this is true iff all $c_j$ are imaginary. Now for imaginary $z = it$, $c_j z^j = c_j i^j t^j$ is real if $j$ is odd and imaginary if $j$ is even.
In order for $f(it)$ to be on the curve $y^2 = x$ (but not at $0$) as (real) $t \to 0$. we would need the first odd $j$ for which $c_j \ne 0$ to be twice the first even $j$ for which $c_j \ne 0$. But this is impossible, because odd $j$ are not twice any integer. So we conclude the only entire function that takes the real axis into the imaginary axis and the imaginary axis into the curve $y^2 = x$ is identically $0$.