Show that an image of a Schlicht function contains ##\Delta(0,1/2)##

[Westlife5]

Homework Statement

Let $f:\Delta\to \mathbb{C}$ be an injective holomorphic function such that $f(0)=0$ and $f'(0)=1$

(i.e $f\in\, S$ is schlicht).

Assume that $A=f(\Delta)$ is convex and let $r\in(0,1)$ and $e^{i\theta}\in\partial\Delta$


$1$ Prove that

$$\frac{1}{2}re^{i\theta}=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz $$


$2$ using part #1#, show that

$$\frac{1}{2}re^{i\theta}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(re^{ix})\cos^2(\frac{\theta-x}{2})dx$$


$3$Show that $\frac{1}{2}re^{i\theta}\in A$

$hint$: use that $\int_{-\pi}^{\pi}\cos^2(\frac{\theta}{2})=\pi$ and that $A$ is convex. In particular use the fact that given $z_0\not\in A$, there exsits $z\in\mathbb{C}$ such that $\langle z,z_0\rangle >\langle z,w \rangle$ for every $w\in A$, where $\langle z,w\rangle$ denotes the standard scalar product between $w,z\in\mathbb{C}$ seen as elements of $\mathbb{R}^2$

Relevant Equations

Not sure what to write here

Hello everyone this was a problem on one of the exams from last year and I'm having trouble with the last point $3$

my solution for $1$
$$\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz =$$
I divided this integral into 3 different ones and got.
$$\int_{|z|=r}\frac{f(z)}{z}dz=f(0)=0$$
$$\int_{|z|=r}\frac{f(z)}{2re^{i\theta}}=\frac{1}{2re^{i\theta}}\int_{|z|=r}f(z)dz=0$$ since f is a holomorphic function
$$\frac{re^{i\theta}}{2}\int_{|z|=r}\frac{f(z)}{z^2}dz=\frac{re^{i\theta}}{2}f'(0)2\pi i=\frac{re^{i\theta}}{2}2\pi i$$so if we put everything together we get

$$\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz =\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^2}=\frac{1}{2\pi i}re^{i\theta}\pi i=\frac{1}{2}re^{i\theta}$$

my solution for $2$
First I parametrized the curve $|z|=r$ by setting $z=re^{i\psi}$ therefore $dz=re^{i\psi}id\psi$ substituting this into the equation I get
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+\frac{1}{2}e^{i(\psi-\theta)}+\frac{1}{2}e^{i(\theta-\psi}))d\psi=
$$
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+\frac{\cos(\psi-\theta)+\cos(\theta-\psi)}{2}+i\frac{\sin(\theta-\psi)+\sin(\psi-\theta)}{2}d\psi=
$$
sines cancel, cosines add up+ half angle formula
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+cos(\psi-\theta))d\psi=\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})2\cos^2(\frac{\theta-\psi}{2})d\psi
$$
which is what we wanted.

$3$

This is the one I'm having trouble with. I know that I have to write $\frac{1}{2}re^{i\theta}$ as an element of the image $f(z)=\frac{1}{2}re^{i\theta}$ for some z. I've though about using something like the Cauchy mean value theorem, however as far as I'm aware that theorem can only be used if we are mapping from $\mathbb{R}\to\mathbb{R}$ (unless it can be used in this case too?). But I'm unsure of how to use the hint.
Could somebody give me a hint of how to use the hint?

 

[lippyka]

The hint can be used to construct a function $g(z)$ such that $f(z)=\frac{1}{2}re^{i\theta}$ if and only if $g(z)=0$. You can use the fact that $f(z)-\frac{1}{2}re^{i\theta}$ is holomorphic, so by the Cauchy mean value theorem, there exists a point $z_0$ such that $$f(z_0)-\frac{1}{2}re^{i\theta}=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)-\frac{1}{2}re^{i\theta}}{z-z_0}dz=0$$Therefore, let $g(z)=f(z)-\frac{1}{2}re^{i\theta}$, then you can solve for $z_0$ to get your desired result.