How to prove that $\frac{||z|-|a||}{1-|a||z|}\leqslant\left|\frac{z-a}{1-\bar{a}z}\right|\leqslant\frac{|z|+|a|}{1+|a||z|}$?

Question

Let $a,z$ be complex numbers such that $|z|<1,|a|<1$, how to prove

$$\dfrac{||z|-|a||}{1-|a||z|}\leqslant\left|\dfrac{z-a}{1-\bar{a}z}\right|\leqslant\dfrac{|z|+|a|}{1+|a||z|}?$$

I've been thinking about it for a while, but I can not work it out.

Answer 1

The second inequality is equivalent to $$|z-a|^2(1+|a||z|)^2\leq |1-\bar{a}z|^2(|z|+|a|)^2$$ that is $$(|z|^2-2\text{Re}(\bar{a}z)+|a|^2)(1+|a||z|)^2\leq (1-2\text{Re}(\bar{a}z)+|a|^2|z|^2)(|z|+|a|)^2$$ The LHS is $$|z|^2-2\text{Re}(\bar{a}z)+|a|^2+2|a||z|^3-4|a||z|\text{Re}(\bar{a}z)+2|a|^3|z|+|a|^2|z|^4-2|a|^2|z|^2\text{Re}(\bar{a}z)+|a|^4|z|^2.$$ The RHS is $$|z|^2-2|z|^2\text{Re}(\bar{a}z)+|a|^2|z|^4 +2|a||z|-4|a||z|\text{Re}(\bar{a}z)+2|a|^3|z|^3 +|a|^2-2|a|^2\text{Re}(\bar{a}z)+|a|^4|z|^2.$$ The RHS minus the LHS gives $$2 (1-|a|^2) (1-|z|^2)(|a||z| + \text{Re}(\bar{a}z))$$ which is non-negative because $|z|<1$, $|a|<1$, and $$|\text{Re}(\bar{a}z))|\leq |\bar{a}z|= |a||z|\implies |a||z| + \text{Re}(\bar{a}z)\geq 0.$$ A similar approach works for the first inequality.

Answer 2

Directly,$$\begin{aligned} &\left|\dfrac{z-a}{1-\bar{a}z}\right |^{2}-\dfrac{(|z|-|a|)^{2}}{(1-|a||z|)^{2}}=\dfrac{(z-a)(\bar{z}-\bar{a})}{(1-\bar{a}z)(1-a\bar{z})}-\dfrac{|z|^{2}+|a|^{2}-2|z||a|}{(1-|a||z|)^{2}}\\ &=\dfrac{(|z|^{2}+|a|^{2}-z\bar{a}-a\bar{z})(1+|a|^{2}|z|^{2}-2|a||z|)-(1+|a|^{2}|z|^{2}-a\bar{z}-\bar{a}z)(|z|^{2}+|a|^{2}-2|z||a|)}{(1-\bar{a}z)(1-a\bar{z})(1-|a||z|)^{2}}\\ &=\dfrac{2|a||z|(1+|a|^{2}|z|^{2}-|z|^{2}-|a|^{2})-(a\bar{z}+\bar{a}z)(1+|a|^{2}|z|^{2}-|z|^{2}-|a|^{2})}{(1-\bar{a}z)(1-a\bar{z})(1-|a||z|)^{2}}\\ &=\dfrac{2(1+|a|^{2}|z|^{2}-|z|^{2}-|a|^{2})(|a||z|-\mathrm{Re}(\bar{a}z))}{(1-\bar{a}z)(1-a\bar{z})(1-|a||z|)^{2}}\\ &=\dfrac{2(1-|z|^{2})(1-|a|^{2})(|a||z|-\mathrm{Re}(\bar{a}z))}{(1-\bar{a}z)(1-a\bar{z})(1-|a||z|)^{2}}\geq0 \end{aligned}$$ since $|a|,|z|<1$ and $|az|\geq|\mathrm{Re}(\bar{a}z)|.$

Similarly,$$\begin{aligned} &\dfrac{(|z|+|a|)^{2}}{(1+|a||z|)^{2}}-\left|\dfrac{z-a}{1-\bar{a}z}\right |^{2}=\dfrac{|z|^{2}+|a|^{2}+2|z||a|}{(1+|a||z|)^{2}}-\dfrac{(z-a)(\bar{z}-\bar{a})}{(1-\bar{a}z)(1-a\bar{z})}\\ &=\dfrac{(|z|^{2}+|a|^{2}+2|z||a|)(1+|a|^{2}|z|^{2}-a\bar{z}-\bar{a}z)-(|z|^{2}+|a|^{2}-a\bar{z}-\bar{a}z)(1+|a|^{2}|z|^{2}+2|a||z|)}{(1+|a||z|)^{2}(1-\bar{a}z)(1-a\bar{z})}\\ &=\dfrac{2|a||z|(1+|a|^{2}|z|^{2}-|z|^{2}-|a|^{2})+(a\bar{z}+\bar{a}z)(1+|a|^{2}|z|^{2}-|a|^{2}-|z|^{2})}{(1+|a||z|)^{2}(1-\bar{a}z)(1-a\bar{z})}\\ &=\dfrac{2(1+|a|^{2}|z|^{2}-|z|^{2}-|a|^{2})(|a||z|+\mathrm{Re}(\bar{a}z))}{(1+|a||z|)^{2}(1-\bar{a}z)(1-a\bar{z})}\\ &=\dfrac{2(1-|z|^{2})(1-|a|^{2})(|a||z|+\mathrm{Re}(\bar{a}z))}{(1+|a||z|)^{2}(1-\bar{a}z)(1-a\bar{z})}\geq0. \end{aligned}$$

Hence we conclude$$\dfrac{\left ||z|-|a|\right |}{1-|a||z|}\leq\left |\dfrac{z-a}{1-\bar{a}z}\right |\leq\dfrac{|z|+|a|}{1+|a||z|}.$$