I'm trying to probe that the region $|z-z_0|\leq R$ maps conformally onto the unit circle $|w|\leq 1$ under the bilinear transformation: $$w=f(z)=\frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda}$$ Where $z_0$ is the center of the circle in the $z$ plane, and $\alpha$ is a point that maps onto the origin of the $w$ plane (i.e., $w=f(\alpha)=0$).
I'm following a book proof (the question is in the last lines at the end):
Let $$w=\frac{az+b}{cz+d} \tag{1}$$ with $ad-bc \neq 0$, be the required transformation. The points $w=0$ and $w=\infty$ are the images of $z=-b/a$ and $z=-d-c$ respectively. Since $0$ and $\infty$ are inverse (or symmetric) points with respect to the circle $|w|=1$, the points $-b/a$ and $-c/d$ , must be inverse (symmetric) points with respect to the circle $|z-z_0|=R$. Now, the inverse point of the point $z=\alpha$ with respect to the circle $|z-z_0|=R$ is $$\alpha^*=z_0+\frac{R^2}{\overline\alpha-\overline z_0}$$
Thus we can write: $\alpha = -\frac{b}{a}$ and $\alpha^* = -\frac{d}{c}= z_0+\frac{R^2}{\overline\alpha-\overline z_0}$ This way, the transformation $(1)$ reduces to (some steps omitted): $$w=\frac{a(\overline\alpha-\overline z_0)}{c} \frac{z-\alpha}{(z-z_0)(\overline\alpha-\overline z_0)-R^2}$$ Now, by applying the condition that any point on $|z-z_0|=R$ must correspond to a point on $|w|=1$, which means that $|w|=1$ when $|z-z_0|=R$ or $(z-z_0)(\overline z-\overline z_0)=R^2$ we have: $$ 1 =|w| =\left|\frac{a(\overline\alpha-\overline z_0)}{c} \frac{z-\alpha}{(z-z_0)(\overline\alpha-\overline z_0)-R^2} \right| = \left|\frac{a(\overline\alpha-\overline z_0)}{c} \right| \left| \frac{z-\alpha}{(z-z_0)(\overline\alpha-\overline z_0)-(z-z_0)(\overline z-\overline z_0)} \right| $$ $$ 1 =\left|\frac{a(\overline\alpha-\overline z_0)}{c} \right| \left| \frac{z-\alpha}{(z-z_0)(\overline\alpha-\overline z)} \right| = \left|\frac{a(\overline\alpha-\overline z_0)}{c} \right| \left| \frac{1}{(z-z_0)} \right|=\left|\frac{a(\overline\alpha-\overline z_0)}{c} \right| \frac{1}{R} $$ Therefore $$\left|\frac{a(\overline\alpha-\overline z_0)}{c} \right| = R $$ and then: $$\frac{a(\overline\alpha-\overline z_0)}{c} = Re^{i\mu} $$ for some real value of $\mu$. So the required transformation is $$w=\frac{R(z-\alpha)}{(z-z_0)(\overline\alpha-\overline z_0)-R^2}e^{i\mu}$$ By taking $\mu = \lambda + \pi$ with $\lambda \in \mathbb{R}$: $$w=\frac{R(z-\alpha)}{(z-z_0)(\overline\alpha-\overline z_0)-R^2}e^{i\lambda}e^{i\pi}=\frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \tag{2}$$ which has the required form. Now, it's clear that $\alpha$ will map onto $w=0$, and we have to verify that the interior of both circles (in $z$ plane and $w$ plane) corresponds to each other. So we have to show that $|w|<1$ when $|z-z_0|<R$, or, $(z-z_0)(\overline z-\overline z_0)<R^2$. Here: $$|w|=\left| \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \right| = \left| \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)} \right| < \left| \frac{R(z-\alpha)}{(z-z_0)(\overline z-\overline z_0)-(z-z_0)(\overline\alpha-\overline z_0)} \right|$$ $$\therefore |w| < \left| \frac{R(z-\alpha)}{(z-z_0)(\overline z-\overline \alpha)} \right| = \frac{|R(z-\alpha)|}{|(z-z_0)(\overline z-\overline \alpha)|} = \color{red}{\frac{R}{|(z-z_0)|}\underset{???}{>}1}$$
Because $|z-z_0|<R \Rightarrow 1<\frac{R}{|z-z_0|}$.
Did I do something wrong? Can I argue something to come up with $|w|<1$ somehow, as required? Is this proof flawed at some point? The proof on the book stops here and, actually states that $\frac{R}{|z-z_0|}<1$, which is wrong as far as my knowledge let me to see.
My another attempt:
Given that $|w| \leq 1$ under such transformation, we can also argue that $|w|^2-1 \leq 0$ or $w \overline w-1 \leq 0$ for $|z-z_0|<R$. This way:
$$\left( \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \right)\overline{\left( \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \right)}-1$$ will lead to:
$$\left( \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \right)\overline{\left( \frac{R(z-\alpha)}{R^2-(z-z_0)(\overline\alpha-\overline z_0)}e^{i\lambda} \right)}-1 \leq 0$$
and ultimately to:
$$R^2|z-\alpha|^2-|R^2-(z-z_0)(\overline{\alpha}-\overline{z_0})|^2 \leq 0$$
My mind is numb and foggy at this point and I can't see how to proceed with this. I will appreciate some help here.
Yet another attempt.
Considering that $\alpha$ is an interior point of the disc $|z-z_0|\leq R$ we have that $|z_0-\alpha|<R$. This is justified because there's no restriction on where the $\alpha$ point can be placed as long we can construct its symmetric point with respect to the circle $|z-z_0|= R$. And because $\alpha$ will be mapped into $w=0$ under $(2)$, so an interior point of $|z-z_0|\leq R$ is being mapped into another interior point of $|w|\leq 1$.
Then, the center $z_0$ of $|z-z_0|= R$ is mapped to $$w_0=\frac{R(z_0-\alpha)e^{i\lambda}}{R^2-(z_0-z_0)(\overline{\alpha}-\overline{z_0})}=\frac{R(z_0-\alpha)e^{i\lambda}}{R^2}=\frac{(z_0-\alpha)e^{i\lambda}}{R}$$ $$\therefore |w_0|=\frac{|z_0-\alpha|}{R} < \frac{R}{R}=1$$
Now, given that $\alpha$ can be any interior point of $|z-z_0|\leq R$, the correspondence between the interiors of both circles on $z$ and $w$ planes is verified.