I have a maximum modulus principle exercise question and I'm stuck trying to understand the solution at the moment.
Here goes:
Let $c\in \mathbb{D}= \left\{z \in\mathbb{C}: |z|<1\right\}$ and define $f:\overline{\mathbb{D}} \mapsto \mathbb{C}$ by
$$f(z)=\frac{z-c}{1-\overline{c}z}: (z \in \overline{\mathbb{D}})$$
Show that $|f(z)|=1$ for $|z|=1$
Hence show that... (I've ommitted this as it's not related to the issue I'm having).
Now in the solutions, I read:
Let $z \in\delta\mathbb{D}$, then $z=e^{i\theta}$ for some $\theta$ and $$f(z)=\frac{e{i\theta}-c}{1-\overline{c}e^{i\theta}}=e^{i\theta}\frac{1-ce^{-i\theta}}{1-\overline{c}e^{i\theta}}$$
Hence $|f(z)|=1$
I do not understand this final step, as I cannot see how the absolute value of the fraction part of the above term is equal to 1, although I understand why $|e^{i\theta}|$ would.
$1-\overline{c}e^{i\theta}$ is the complex-conjugate of $1-ce^{-i\theta}$ and a complex number ant its conjugate obviously have the same module, being one the symmetric of the other with respect to the real axis.
EDIT: properties of complex conjugation: for every $z,w\in\mathbb{C}$,
1) $\overline{z+w}=\overline{z}+\overline{w}$
2)$\overline{z\cdot w}=\overline{z}\cdot\overline{w}$
In you case, you have $$\overline{1-ce^{-i\theta}}=\overline{1}-\overline{c}\overline{e^{-i\theta}}$$ You can conclude noticing that $\overline{1}=1$ and $\overline{e^{-i\theta}}=\overline{cos\theta-i\sin\theta}=\cos\theta+i\sin\theta=e^{i\theta}$