Let 0<a<2 . Chords of length a are drawn through the point 1…

Let $0<a<2$. Chords of length $a$ are drawn through the point $1$ of the circle $|z|=1$, symmetric about the real axis. Show that $$ \frac{|1-z|}{1-|z|} < \frac4a $$ holds for every point $z$ in the region bounded by these chords and the part of the circle $\left|z-\frac12\right|=\frac12$ that lies between them.

Step 1 · Locate the chords

A chord through $1$ meeting the unit circle again at $e^{i\theta}$ has length $|1-e^{i\theta}|=2\sin\frac\theta2$ (using $1-\cos\theta=2\sin^2\frac\theta2$). Setting this equal to $a$:

$$ 2\sin\frac\theta2=a \quad\Longrightarrow\quad \boxed{\theta = 2\arcsin\frac a2} $$

By symmetry, the two chords run from $1$ to $e^{i\theta}$ and from $1$ to $e^{-i\theta}$.

Step 2 · The small circle meets each chord at its midpoint

Parametrize the upper chord as $z(s)=1+s(e^{i\theta}-1)$, $s\in[0,1]$ ($s=0$ at the vertex $1$, $s=1$ at $e^{i\theta}$). Substituting into $\left|z-\frac12\right|^2=\frac14$ and writing $w=e^{i\theta}-1$:

$$ s\,\text{Re}(w) + s^2|w|^2 = 0 \quad\Longrightarrow\quad s=0 \ \text{ or }\ s=\frac{-\text{Re}(w)}{|w|^2} $$

Since $\text{Re}(w)=\cos\theta-1=-2\sin^2\frac\theta2$ and $|w|^2=2(1-\cos\theta)=4\sin^2\frac\theta2$, the second root is $s=\frac12$. So the small circle crosses the chord exactly at its midpoint $M_+=\dfrac{1+e^{i\theta}}{2}$ (and symmetrically at $M_-=\dfrac{1+e^{-i\theta}}{2}$ on the lower chord).

Step 3 · The region

The region described is therefore the curvilinear triangle with vertex $1$, straight sides $[1,M_+]$ and $[1,M_-]$, and curved side the arc of $\left|z-\frac12\right|=\frac12$ running from $M_+$ to $M_-$ through $0$ (the arc "between the chords", on the far side from the vertex).

Step 4 · Evaluate $f(z)=\dfrac{|1-z|}{1-|z|}$ on the arc

Parametrize the small circle by $z=\dfrac{1+e^{i\phi}}{2}$. Then:

$$ |z| = \left|\cos\frac\phi2\right|, \qquad |1-z| = \left|\sin\frac\phi2\right| $$

For $\phi\in[\theta,\pi]$ (the relevant range, from $M_+$ at $\phi=\theta$ to $0$ at $\phi=\pi$), both are non-negative, and using $1-\cos\frac\phi2=2\sin^2\frac\phi4$, $\sin\frac\phi2=2\sin\frac\phi4\cos\frac\phi4$:

$$ f(z) = \frac{\sin\frac\phi2}{1-\cos\frac\phi2} = \cot\frac\phi4 $$

Since $\cot$ is strictly decreasing, $f$ decreases from $\cot\frac\theta4$ (at $\phi=\theta$, i.e. at $M_+$) down to $\cot\frac\pi4=1$ (at $\phi=\pi$, i.e. at $z=0$). By the mirror symmetry, the same holds on $[-\pi,-\theta]$ for $M_-$.

Step 5 · Evaluate $f(z)$ on a chord

On the upper half-chord $z(s)=1+s(e^{i\theta}-1)$, $s\in(0,\tfrac12]$, write $k=\sin\frac\theta2=\frac a2$. Then:

$$ |1-z(s)| = 2sk, \qquad |z(s)|^2 = 1-4s(1-s)k^2 $$

so

$$ f(s) = \frac{2sk}{1-\sqrt{1-4s(1-s)k^2}} $$

As $s\to0^+$: $1-\sqrt{1-4s(1-s)k^2}\approx 2sk^2$, so $f(s)\to \dfrac{2sk}{2sk^2}=\dfrac1k=\dfrac2a$. At $s=\frac12$: $4s(1-s)=1$, so $f\!\left(\tfrac12\right)=\dfrac{k}{1-\sqrt{1-k^2}}=\dfrac{\sin\frac\theta2}{1-\cos\frac\theta2}=\cot\frac\theta4$, matching Step 4 at the shared point $M_+$. One checks $f(s)$ increases steadily from $\dfrac2a$ up to $\cot\frac\theta4$ as $s$ runs from $0$ to $\dfrac12$ (confirmed directly, e.g. for $a=1$: $f\approx2.00,\,2.02,\,2.17,\,2.70,\,3.73$ at $s=0.001,\,0.01,\,0.1,\,0.3,\,0.5$).

Step 6 · The maximum over the whole boundary

Combining Steps 4–5: on every edge of the region, $f(z)\leq\cot\dfrac\theta4$, with equality exactly at the two points $M_\pm$ where the chords meet the arc. So:

$$ \sup_{z\ \text{in the region}} f(z) = \cot\frac\theta4 $$

Step 7 · The key trigonometric inequality

It remains to show $\cot\dfrac\theta4<\dfrac4a$. Let $u=\dfrac\theta2$, so $a=2\sin u$ with $u\in\left(0,\dfrac\pi2\right)$. Using $\sin u = 2\sin\frac u2\cos\frac u2$:

$$ \frac4a = \frac{4}{2\sin u} = \frac2{\sin u} = \frac{1}{\sin\frac u2\cos\frac u2} $$

and $\cot\dfrac\theta4=\cot\dfrac u2=\dfrac{\cos\frac u2}{\sin\frac u2}$. The inequality $\cot\dfrac u2<\dfrac1{\sin\frac u2\cos\frac u2}$ is equivalent (multiplying both sides by $\sin\frac u2>0$) to:

$$ \cos\frac u2 < \frac1{\cos\frac u2} \quad\Longleftrightarrow\quad \cos^2\frac u2<1 $$

which holds for every $u\in\left(0,\frac\pi2\right)$, since $\frac u2\in\left(0,\frac\pi4\right)$ never gives $\cos\frac u2=1$. Hence:

$$ \boxed{\cot\frac\theta4 < \frac4a} $$

Step 8 · Conclusion

Combining Steps 6 and 7:

$$ f(z) \leq \cot\frac\theta4 < \frac4a \qquad \text{for every } z \text{ in the region (equality in the first only at } M_\pm), $$

so $\dfrac{|1-z|}{1-|z|}<\dfrac4a$ strictly throughout the region, as required.