Mapping of $f(z)$ Let the function $f$ be analytic in the complex plane, real on real axis, 0 at origin and not identically zero.
Prove that, if $f$ maps imaginary axis into a straight line, then, that straight line must be either real axis or imaginary axis.
My effort: $f(z)$ is analytic iff $g(z)= \overline{f(\bar z)}$ is also analytic.$f(z)$ coincide with $g(z)$ on real axis. Consider the sequence ${1/n}$ converges to zero. Now, using identity theorem we can conclude $f(z)=g(z)$ over complex plane. $g(z)$ maps imaginary axis to imaginary axis and so is $f(z)$. I can not understand when $f$ maps imaginary axis to real axis.
 A: Let $k \ge 1$ the order of the zero of $f$ at the origin; by the local form of the analytic function at $0$, namely, $f(z)=cz^k+O(z^{k+1}), c \ne 0$, it immediately follows that $f$ transforms the angle $\theta$ between any two curves passing through the origin, to the angle $k\theta$ (in particular $f$ is conformal at the origin precisely for $k=1$)
Since the angle between the real and imaginary axis is $\pi/2$, the angle between their images is $k\pi/2$, so by hypothesis, the imaginary axis is sent into a line that makes a $k\pi/2$ angle with the real axis for some integer $k \ge 1$ and there are only two such lines, the imaginary and real axis depending whether $k$ is odd or even, so we are done.
$z^2, z^4$ are examples that satisfy the hypothesis and where the imaginary axis is sent to the real axis (though in one case the two images are disjoint except at zero being two half-lines and in the other, they coincide - note that the image of the real or imaginary axis under $f$ may not be onto the full line it gets sent too), while $z$ is an example where it is sent to itself
