Improvement of the Koebe 1/4 theorem for functions with rotational symmetry Reading this question I don't get where the hypothesis of $ f(\mathbb{D}) $ convex it is used to prove the statement.
Sharp radius for univalent convex functions
I suppose that if $ f(\mathbb{D})$ is not convex then
$$ h (z) = \frac{w^2 - (f(z)-w)^2}{2w} $$
is not injective but I don't see why.I'm asking this because I'm trying to solve a similar exercice that say:
Suppose that $ f \in \mathcal{S} $ with $f(iz)=if(z) $ for every $z \in \mathbb{D} $. Show that $f(\mathbb{D}) $ contain $B(0,1/\sqrt{2}) $. My idea it is actually to do the a similar argumentation constructing the function
$$ h(z)= \frac{ w^{2\sqrt{2}} - (w -f(z))^{2 \sqrt{2}} }{2\sqrt{2} w^{2\sqrt{2}-1}} $$
In fact we have that if $w \not\in f(\mathbb{D})$ then $\frac{w}{2\sqrt{2}} \not\in h(\mathbb{D}) $ and thus concluding with 1/4 Koebe theorem that $ \left| \frac{w}{2\sqrt{2}} \right| \geq 1/4 $ and thus that $ \left| w \right| \geq 1/\sqrt{2} $. So I have to prove that $ h $ is actually in $\mathcal{S} $ when $f \in \mathcal{S} $ satisying $f(zi)=if(z)$ but I don't see how.
If $f(0)=0$ and $f'(0)=1$ we have that $ h(0)=0$ and $h'(0)=1$. We have that $f$ is holomorphic so $h$. So the condition $f(iz)=if(z)$ has to be necessary to prove that $h$ is injective. Same doubt with the original problem the hypothesis that $f(\mathbb{D})$ is convex has to be necessary to prove that $h$ is injective. But why?
 A: Regarding the first part of your question:
If $f$ is injective, $w \notin f(\Bbb D)$ and $f(\Bbb D)$ is convex then $(f(z)-w)^2$ is injective. The reason is that
$$
 (f(z_1)-w)^2 = (f(z_2)-w)^2 \implies f(z_1) = f(z_2) \implies z_1 = z_2 
$$
because
$$ (f(z_1) - w) = -(f(z_2) - w) \implies w = \frac 12 (f(z_1) + f(z_2))
$$
is not possible. It follows that $h$ is injective.
This argument no longer works if $f(\Bbb D)$ is not convex. As a concrete counterexample, we can choose the Koebe function $k(z) = z/(1-z)^2$ and $w = -1/4$. Then
$$
 h(z) = -\frac 18 + \frac 18 \left(\frac{z+1}{z-1}\right)^4
$$
is not injective, for example
$$
 h(\frac 15 + \frac 25 i) = -\frac 58 = h(\frac 15 - \frac 25 i) \, .
$$

Regarding the second part of your question:
I don't think that your actual problem is related to the Koebe quarter theorem for convex functions or can be proved in a similar way. Instead one can proceed as follows:
If $f \in \cal S$ with $f(iz) = if(z)$ then $f(z)^4 = h(z^4)$ for some function $h \in \cal S$. The Koebe quarter theorem applied to $h$ states that $h(\Bbb D)$ contains the disk $B(0,1/4)$, which implies that $f(\Bbb D)$ contains the disk $B(0, 1/\sqrt 2)$.
To show the existence of $h$ we note that
$$
 f(z) = z + a_5 z^5 + a_9 z^9 + \cdots = z g(z^4)
$$
where $g$ is holomorphic in $\Bbb D$ with $g(0) = 1$, and define
$$
 h(w) = w g(w)^4 \, .
$$
$h$ is is holomorphic in $\Bbb D$ with $h(0) = 0$, $h'(0) = 1$, and $h(z^4) = f(z)^4$.
It remains to show that $h$ is injective: Assume that $h(w_1) = h(w_2)$ with $w_1, w_2 \in \Bbb D$. Choose $z_1, z_2 \in \Bbb D$ with $z_1^4 = w_1$ and $z_2^4 = w_2$. Then
$$
 f(z_1)^4 = g(w_1) = g(w_2) = f(z_2)^4
$$
so that
$$
 f(z_1) = i^k f(z_2) = f(i^k z_2)
$$
for some $k \in \{ 0, 1, 2, 3 \}$. Since $f$ is injective it follows that $z_1 = i^k z_2$ and consequently $w_1 = w_2$. This concludes the proof.

In the same way one can show that if $f \in \cal S$ satisfies $f(\omega z) = \omega f(z)$ where $\omega \ne 1$ is a primitive $n$-th root of unity then $f(\Bbb D)$ contains the disk $B(0, 1/\sqrt[n]{4})$.
