# Clarification over Ahlfors page 116, 2.1 about winding numbers

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.

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 $$b$$.

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.



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.

• You seem to have missed the "$\leq 0$" part of the statement. – Eric Wofsey Oct 27 '18 at 3:32

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 $$b$$.