PROBLEM STATEMNT
Let $f:\Delta\to \mathbb{C}$ be an injective holomorphic function such that $f(0)=0$ and $f'(0)=1$
(i.e $f\in\, S$ i¸s 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 $\mathbb{R}^2$
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 ATTEMPT
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?
Note that part 2 states that $\tfrac12 r \exp(\mathrm i \theta)$ is a convex combination of function values $f(r \exp(\mathrm i x))$. Also note that the bound is sharp since it is attained by $$ f(z) = \frac{z}{1-z}$$ for which $f(-1) = -\tfrac12$.