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Given two points $z_1,z_2$ such that $ \lvert z_i\rvert<1$, show that for every point $z\ne 1$ in the closed triangle with vertices $z_1,z_2,1$ following holds: $$ \frac{\lvert 1-z\rvert}{1-\lvert z\rvert}\le K,$$ where $K$ is a constant that depends only on $z_1, z_2.$ Determine the smallest value of $K$ for $z_1= \frac{1+i}{2}, z_2=\frac{1-i}{2}$.

What I tried, it's to write $z=re^{i\theta}$, then $r<1$, I'll prove the result but for $\left(\dfrac{\lvert 1-z\rvert}{1-\lvert z\rvert}\right)^2$ $= \dfrac{1-2r\cos\theta+r^2}{1-2r+r^2}$, and $\theta$ is bounded by the angles of $z_i$ but I can't see, what I can do now.

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up vote 2 down vote accepted

Hint: It's better to write $z = 1 - r e^{i\theta}$. Then $|1 - z| = r$, while $|z|$ can be obtained using the Law of Cosines.

EDIT. Well, I might as well summarize the whole thing now. Use polar coordinates centred on $1$ rather than $0$, so $z = 1 - r e^{i\theta}$. The two points $z_1$ and $z_2$, and therefore also the triangle, are contained inside some sector $-\pi/2 < -\theta_1 \le \theta \le \theta_1 < \pi/2$, and also inside some smaller circle $r = 2 p \cos \theta$ of radius $p < 1$ with centre on the real axis and passing through $1$, as in the picture below:

enter image description here

For all $z\ne 1$ in the shaded region, and in particular for points in the triangle, $$ \frac{|1-z|}{1-|z|} = \frac{r}{1-\sqrt{1 + r^2 - 2 r \cos \theta}} = \frac{1}{2 \cos \theta - r} \left(1 + \sqrt{1+r^2 - 2 r \cos \theta}\right) $$

Now $2 \cos \theta - r = (2 - 2p) \cos \theta + 2 p \cos \theta - r \ge (2 - 2 p) \cos \theta_1 > 0$, while $1 + \sqrt{1 + r^2 - 2 r \cos \theta} = 1 + |z| \le 2$, so we get $$ \frac{|1-z|}{1-|z|} \le \frac{1}{(1-p) \cos \theta_1}$$

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I don't know How I can bound $\theta$ , specially in the case $z= 1-re^{i\theta}$ sorry for being so stupid!! – Daniel Sep 3 '12 at 0:01
There will be two values of $\theta$ for the two points $z_1, z_2$, and $\theta$ will be between those for every point of the triangle. – Robert Israel Sep 3 '12 at 0:13
Yes, but this is true when I take $z=re^{i\theta}$ in your change of variables, I don't know how to bound $\theta$ (the relation between $r_1e^{i\theta_1} \to 1-r_2e^{i\theta_2}$ it's not trivial – Daniel Sep 3 '12 at 4:01
Don't take $z = r e^{i\theta}$, take $z = 1 - r e^{i\theta}$. $z_1$ is some point in the open unit disk. It is $1 - r_1 e^{i\theta_1}$ for some $r_1$ and $\theta_1$ with $-\pi/2 < \theta_1 < \pi/2$. Note that $r_1 < 2 \cos(\theta_1)$ (points $z_1$ with $r_1 = 2 \cos(\theta_1)$ would be on the circle, since $-1$, $1$ and $z_1$ would form a right triangle). – Robert Israel Sep 3 '12 at 4:50
@Daniel Another way to look at this: The line segment at angle $\theta$ connecting $(1,0)$ to the circle $|z| = 1$ has length $2\cos(\theta)$. But each of the other two vertices of the triangle isn't all the way over to the circle; it is in the interior of the disk. So there's some $\delta > 0$ such that $2\cos(\theta) - r > \delta$ at this point. Similarly, a compactness argument shows that there's a $\delta > 0$ such that $2\cos(\theta) - r > \delta$ at all points in the triangle since the whole triangle is in the interior of the disk. – Zarrax Sep 3 '12 at 21:07

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