Image under conformal mapping I'm trying to prove that the mapping $$f(z) = \frac{z ( az+1)}{z+a}$$ where $a=\sin(\alpha/4)$ ($0< \alpha < 2\pi$) maps $\mathbb{C}_{\infty} \setminus \overline{B}(0;1)$ conformally onto $\mathbb{C}_{\infty}\setminus K$, where $K$ is the circular arc $\left\{ e^{i\theta } : -\alpha/2 \leq \theta \leq \alpha /2 \right\}$ and $\overline{B}(0;1)$ is the closed unit ball. 
I have tried to write $f$ as a composition of analytic functions, i.e $f= f_4 \circ f_3 \circ f_2 \circ f_1$ where $f_1 (z) = z+a$, $f_2 (z) = z+ \frac{a^2-1}{z}, f_3 (z) = z-a + \frac{1-a^2 }{a}$ and $f_4 (z) = az$. But I can't find the image of the set $\mathbb{C}_{\infty} \setminus \overline{B}(0;1) +a$ under the mapping $f_2$. 
Any help would be appreciated. Regards.
 A: I think it would be easier if you split $f_2 = k\circ h\circ g$ with $g(z) = \frac{z}{\sqrt{1-a^2}}$, $h(z) = z - \frac{1}{z}$, and $k(z) = \sqrt{1-a^2}\cdot z$. It would still be fiddly, though, since you have a circle with centre $\neq 0$ whose image you need to find.
I believe it is still easier (though admittedly still nontrivial) to exploit the known symmetries the unit circle gives. I have therefore done that:
Since $0 < a < 1$, there is no cancellation in the representation of $f$, and thus $f$ is a rational function of order(1) $2$. That means every value in the Riemann sphere is attained exactly twice, counting multiplicities. Knowing where we want to end up, we check what the unit circle is mapped to, where the derivative of $f$ vanishes, and how $f$ behaves under reflection in the unit circle.
We start with the reflection in the unit circle:
$$f(1/\overline{z}) = \frac{a/\overline{z}+1}{\overline{z}(1/\overline{z}+a)} = \frac{a+\overline{z}}{\overline{z}(1+a\overline{z})} = 1/\overline{f(z)},\tag{1}$$
so points symmetric with respect to the unit circle are mapped to points symmetric with respect to the unit circle.
Next the derivative:
$$f'(z) = \frac{(z+a)(2az+1) - z(az+1)}{(z+a)^2} = \frac{az^2+2a^2z + a}{(z+a)^2},$$
and the zeros of the derivative are the zeros of $z^2 + 2az + 1 = (z+a)^2 + (1-a^2)$, so they are $-a \pm i \sqrt{1-a^2}$, hence lie (as they should) on the unit circle. Using $a = \sin (\alpha/4)$, the zeros are $-\sin (\alpha/4) \pm i \cos (\alpha/4) = \exp\bigl(\pm i\bigl(\frac{\pi}{2}+\frac{\alpha}{4}\bigr)\bigr)$.
Since, as is easily verified, $f$ maps the unit circle into the unit circle, the two arcs determined by these points are mapped to the same arc of the unit circle. We have $f(1) = 1$, and
\begin{align}
f(-a \pm i\sqrt{1-a^2}) &= \frac{(-a\pm i\sqrt{1-a^2})(1-a^2\pm ia\sqrt{1-a^2})}{\pm ia \sqrt{1-a^2}}\\
&= \mp i (-a\pm i\sqrt{1-a^2})(\sqrt{1-a^2}\pm ia)\\
&= (-a\pm i\sqrt{1-a^2})(a \mp i\sqrt{1-a^2})\\
&= -(-a\pm i\sqrt{1-a^2})^2\\
&= -\exp \bigl(\pm i\bigl(\pi + \tfrac{\alpha}{2}\bigr)\bigr)\\
&= \exp \bigl(\pm i\tfrac{\alpha}{2}\bigr),
\end{align}
so indeed the unit circle is mapped to the circular arc $K$. The other arc of the unit circle is the image of the circle $C$ through $-a \pm i\sqrt{1-a^2}$ that intersects the unit circle orthogonally, that is the circle
$$\biggl\lvert z + \frac{1}{a}\biggr\rvert = \frac{\sqrt{1-a^2}}{a}.$$
An elementary calculation shows that
$$\left\lvert f\left(\frac{e^{i\varphi}\sqrt{1-a^2}-1}{a}\right)\right\rvert = 1,$$
so $C$ is indeed mapped to the unit circle. Since the angles are doubled at the zeros of $f'$, $C$ is mapped to the complement of $K$ (plus the two endpoints of the arc) in the unit circle. Since $f$ has order $2$, no other point is mapped to the unit circle.
The two circles $C$ and $C_1 = \{ z : \lvert z\rvert = 1\}$ partition the Riemann sphere into four regions,
\begin{align}
R_1 &= \left\{ z : \lvert z\rvert < 1, \left\lvert z + \frac{1}{a}\right\rvert > \frac{\sqrt{1-a^2}}{a}\right\},\\
R_2 &= \left\{ z : \lvert z\rvert < 1, \left\lvert z + \frac{1}{a}\right\rvert < \frac{\sqrt{1-a^2}}{a}\right\},\\
R_3 &= \left\{ z : \lvert z\rvert > 1, \left\lvert z + \frac{1}{a}\right\rvert > \frac{\sqrt{1-a^2}}{a}\right\},\\
R_4 &= \left\{ z : \lvert z\rvert > 1, \left\lvert z + \frac{1}{a}\right\rvert < \frac{\sqrt{1-a^2}}{a}\right\}.
\end{align}
Reflection in the unit circle maps $R_1$ anti-conformally to $R_3$ and $R_2$ anti-conformally to $R_4$.
Each of these regions is mapped by $f$ to a region in the Riemann sphere bounded by the unit circle $C_1$. Since $0 \in R_1$ and $-a\in R_2$, and $f(0) = 0,\, f(-a) = \infty$, $R_1$ is mapped by $f$ to the (open) unit disk, and $R_2$ to the complement of the closed unit disk. By $(1)$, it then follows that $R_3$ is mapped to the complement of the closed unit disk, and $R_4$ to the (open) unit disk. Since $f$ has order $2$, all these mappings of $R_k$ to the interior resp. exterior of the unit disk are conformal. It follows that $f$ maps the complement of the closed unit disk conformally to the complement of $K$, as desired.

(1) Here, "order" refers to the order of a rational function as the maximum of the degrees of numerator and denominator (in reduced form). Not to be confused with the (exponential) order of an entire function.
