I am reading Kodaira's Complex Analysis, and trying to understand his proof of the uniformization theorem for simply connected Riemann surfaces. Due to what I assume is translation errors, this book (though very excellent) has many misprints. I have spent several hours trying to figure out what Kodaira actually meant in the following (page 331; Chapter 7 section 1, in the paragraph immediately preceding the statement of theorem 7.4):
If $F$ is the segment connecting $\alpha$ and $\beta = \alpha + \delta$, $\delta > 0$, the transformation $w \rightarrow 2/\delta(w - (\alpha + \beta)/z)$ carries $F$ into the segment $[-1,1]$.
The problem I'm having is that the transformation he gives doesn't take $F$ to that segment at all. Also, there is a $z$ in my edition that I can only assume was supposed to be a $2$. I tried messing around with linear fractional transformations to derive the formula myself with no luck, as I am left with one degree of freedom (which leads me to also consider the possibility that the $z$ is in fact supposed to be a $z$).
Here is what I tried:
Since we're considering biholomorphic maps on the Riemann sphere, the transformation we want will be a Möbius transformation of the form $$f(w) = \frac{aw + b}{cw + d}.$$
We want $f(\alpha) = -1$, $f(\beta) = 1$, and points of the form $\alpha + t \delta$ (for $t \in [0,1]$) to lie in the interval $[-1, 1]$. This gives us the equations:
$$a\alpha + b = c\alpha + d\\ a\beta + b = -c\beta - d$$
If we let $(\alpha + \beta)/2 \mapsto 0$, we get an additional equation, except with this relation we get $c = 0$ which to me seems very wrong:
$$a(\alpha + \beta)/2 + b = 0\\ \implies a(\alpha + \beta) + 2b = 0$$
Hence, $$a \alpha + a \beta + 2b = c (\alpha - \beta) = 0 \implies c = 0$$ So $f(w) = (a/d) w + b/d$.
I'm hoping someone with more expertise on the matter can help me figure this out. I'm sure I'm either doing something wrong, or there's something trivial here I'm missing. Thanks.
EDIT: The $z$ in the expression is indeed supposed to be a $2$. After the wonderful observation of Izaak below, it turns out I foolishly assumed the division symbol was suppose to carry over the entire expression. That is, $$f(w) = \frac{2}{\delta(w - (\alpha + \beta)/2)}.$$ This is not the case. The expression is to be interpreted as $\frac{2}{\delta}(w - (\alpha + \beta)/2)$. I might as well carry through the derivation above (in fact, getting $c = 0$ was the correct result).
If $f(w) = (a/d) w + b/d$, then knowing that $f(\alpha) = -1$ and $f(\beta) = 1$, we get: $$a\alpha + b - d = 0\\ a \beta + b + d = 0$$ which we can solve as a system of linear equations to get: $$a = -\frac{2}{\delta} t\\ b = \frac{\alpha + \beta}{\delta} t\\ d = t$$ where we may choose $t$ however we like. To get the expression Kodaira uses, choose $t = -1$. We get $a = 2/\delta$, and $b = -\frac{\alpha + \beta}{\delta} = -\frac{2}{\delta}\frac{\alpha + \beta}{2}$