I need to prove the following statement: $$ |z||\frac{az + b}{z+d}-a| <= M $$ with $a,b,c,z \in \mathbb{C}, |z| \geq 1 + |d|$ and $M\geq 0$. I have reduced this to $$ |z||b-ad| \leq M $$ Also $ad -b \neq 0$. I don't know how to proceed my proof.
2026-04-24 19:43:30.1777059810
Prove that $|z||b-ad| \leq M $
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$$ \begin{array}{lcl} |z|\big|\frac{az+b}{z+d}-a\big| &=& |z|\big|\frac{az+b}{z+d}-\frac{az+ad}{z+d}\big| \\ &=& |z|\big|\frac{b-ad}{z+d}\big| \\ &=& |b-ad|\big|\frac{z}{z+d}\big| \\ &=& |b-ad|\big|1-\frac{d}{z+d}\big| \\ &\leq & |b-ad|\bigg(1+\frac{|d|}{|z+d|}\bigg) \\ &\leq & |b-ad|\bigg(1+\frac{|d|}{|z|-|d|}\bigg) \\ &\leq & |b-ad|\bigg(1+|d|\bigg) \end{array} $$
So you can take $M=|b-ad|\bigg(1+|d|\bigg)$.