Prove: If $z,\alpha \in \mathbb{C}$ with $|z|<1, |\alpha|<1$ then $\dfrac{|z|^2+|\alpha|^2}{1+|\alpha z|^2}<1$
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Here, $|\alpha|^2+|z|^2-|\alpha z|^2=|\alpha|^2+|z|^2-|\alpha|^2|z|^2-1+1=1-(1-|\alpha|^2)(1-|z|^2)$. Now, Since $|\alpha|\lt 1$ and $|z|\lt 1\implies (1-|\alpha|^2)(1-|z|^2)\gt 0\implies 1-(1-|\alpha|^2)(1-|z|^2)\lt 1$ $$\implies |\alpha|^2+|z|^2-|\alpha z|^2\lt 1$$ $$\implies |\alpha|^2+|z|^2\lt|\alpha z|^2 + 1$$ $$\implies \frac{|\alpha|^2+|z|^2}{|\alpha z|^2 + 1}\lt 1$$ inequality remain unchanged on division as $|\alpha z|^2 + 1\gt 0$ |
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Because $0\leq |z|<1$ and $0\leq |\alpha|<1$, we have that $|z|^2<1$ and $|\alpha|^2<1$. Thus $$|z|^2-1<0\qquad\text{ and }\qquad |\alpha|^2-1<0,$$ so that $$0<(|\alpha|^2-1)(|z|^2-1).$$ Multiplying out the right side, $$0<|\alpha|^2|z|^2-|\alpha|^2-|z|^2+1$$ and of course $|\alpha|^2|z|^2=|\alpha z|^2$, so that $$|z|^2+|\alpha|^2<1+|\alpha z|^2.$$ Because $1+|\alpha z|$ is always a positive real number, we can divide both sides of the inequality by it to obtain $$\frac{|z|^2+|\alpha|^2}{1+|\alpha z|^2}<1.$$ |
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$\dfrac{|z|^2+|\alpha|^2}{1+|\alpha z|^2}<1$ iff $|z|^2+|\alpha|^2 < 1+|\alpha z|^2$ as both expressions are > 0 iff $(1-|z|^2)(1-|\alpha|^2)<0$ $(1-|z|^2)<0\ if\ |z|^2,1\ if\ |z|<1 $ |
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