Clarification over Ahlfors page 116, 2.1 about winding numbers

Question

Everything on this question is in complex plane.

As the book describes a property of a winding number, it says that:

Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $\leq 0$.

Here, the above statement should be interpreted as "never (real and $\leq 0$)".

If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $\leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.

Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.

Answer 1

Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.

Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k \le 0$. That is, $$a - kb = (1 - k)c\\ \text{Let } l = (1 - k)^{-1}, \text{where } 0 \lt l \le 1.\\ c = {a - kb\over 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$

Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c \neq a + m(b - a)$ for any real number $0 \le m \le 1$. But we have just found $0 \le 1 - l \lt 1$ above. So the statement can only be false.

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Original answer:

I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.

Answer 2

Note that $\frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=\frac{z}{z-b}.$$ Solving for $z$, we have $$z=\frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $t\leq 0$, then $\frac{t}{t-1}\in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.

From a geometric perspective, $\frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.

Answer 3

The imaginary part of $u=\frac{z-a}{z-b}$ is $$\frac{u-\overline{u}}{2i}=\frac{1}{2i}\left(\frac{z-a}{z-b}-\overline{\frac{z-a}{z-b}}\right)=\frac{1}{2i}\left(\frac{z-a}{z-b}-\frac{\overline{z}-\overline{a}}{\overline{z}-\overline{b}}\right)$$

So, if we want this to be real, we need

$$ \begin{eqnarray*} \frac{1}{2i}\left(\frac{z-a}{z-b}-\frac{\overline{z}-\overline{a}}{\overline{z}-\overline{b}}\right)&=&0 \\ (z-a)(\overline{z}-\overline{b})&=&(z-b)(\overline{z}-\overline{a}) \\ z\overline{z}-z\overline{b}-a\overline{z}+a\overline{b}&=&z\overline{z}-z\overline{a}-b\overline{z}+b\overline{a} \\ 0&=&z(\overline{a}-\overline{b})+b(\overline{z}-\overline{a})-a(\overline{z}-\overline{b}) \\ 0&=&-\Re(b)\Im(z)+\Im(b)\Re(z)+\Re(a)(\Im(z)-\Im(b))+\Im(a)(\Re(b)-\Re(z)) \\ 0&=&\Re(b)\Im{a}-\Re(a)\Im(b)-(\Re(a)-\Re(b))\Im(z)+(\Im(b)-\Im(a))\Re(z) \\ \Im(z)&=&\frac{\Im(b) \Re(a) - \Im(a) \Re(b) + (\Im(a) - \Im(b)) \Re(z)}{\Re(a) - \Re(b)} \end{eqnarray*} $$ which is the complex point-slope form of the line connecting $a$ and $b$.

So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.

$$ \begin{eqnarray*} u&=&\frac{\Re(z)+i\Im(z)-\left(\Re(a)-i \Im(a)\right)}{\Re(z)+i\Im(z)-\left(\Re(b)-i \Im(b)\right)}\\ &=& \frac{\Re(z)+i\left(\frac{\Im(b) \Re(a) - \Im(a) \Re(b) + (\Im(a) - \Im(b)) \Re(z)}{\Re(a) - \Re(b)}\right)-(\Re(a)+i \Im(a))}{\Re(z)+i\left(\frac{\Im(b) \Re(a) - \Im(a) \Re(b) + (\Im(a) - \Im(b)) \Re(z)}{\Re(a) - \Re(b)}\right)-(\Re(b)+i \Im(b))}\\ &=& \frac{\left(\Re(z)-\Re(a)\right)\left(\frac{\Re(a)-\Re(b)+i(-\Im(a)+\Im(b))}{\Re(a)-\Re(b)}\right)}{\left(\Re(z)-\Re(b)\right)\left(\frac{\Re(a)-\Re(b)+i(-\Im(a)+\Im(b))}{\Re(a)-\Re(b)}\right)}\\ &=& \frac{\Re(z)-\Re(a)}{\Re(z)-\Re(b)} \end{eqnarray*} $$ From there, it's simple algebra to see that $u=\frac{\Re(z)-\Re(a)}{\Re(z)-\Re(b)}$ is negative only between $\Re(a)$ and $\Re(b)$, which corresponds to the line segment joining $a$ and $b$.