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Let $z,w\in\mathbb{C}$ and $|z|,|w|<1$. Show that $\displaystyle \left|\frac{z-w}{1-z\bar{w}}\right|<1$.

My try:

$$\left|\frac{z-w}{1-z\bar{w}}\right|^2=\left(\frac{z-w}{1-z\bar{w}}\right)\left(\frac{\bar{z}-\bar{w}}{1-\bar{z}w}\right)=\frac{|z|^2-z\bar{w}-\bar{z}w+|w|^2}{1-\bar{z}w-z\bar{w}+|z|^2|w|^2}=\frac{|z|^2-2\text{Re}(z\bar{w})+|w|^2}{1-2\text{Re}(z\bar{w})+|z|^2|w|^2}$$

I tried to show that the fraction is less than $1$, but I couldn't. How can I continue?

Please help, thanks!

Galc127
  • 4,451

1 Answers1

4

Hint

add $$\Longleftrightarrow \dfrac{|z|^2-2\rm{Re}(z\overline{w})+|w|^2}{1-2\rm{Re}(z\overline{w})+|z|^2|w|^2}<1$$ $$\Longleftrightarrow |z|^2-2\rm{Re}(z\overline{w})+|w|^2<1-2\rm{Re}(z\overline{w})+|z|^2|w|^2$$ $$1-|w|^2-|z|^2+|w|^2|z|^2>0$$ $$\Longleftrightarrow (1-|w|^2)(1-|z|^2)> 0$$

math110
  • 93,304