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Let $\alpha$ be a fixed complex number such that $ |\alpha| $ < 1 and $$ w = \frac{z-\alpha}{1-\alpha \bar z} $$ where z is a complex number.

Prove that $|w|<1 $ for all $z$ such that $|z|<1$.

Attempt. Putting $z$ as $x+iy$ and $\alpha$ as some $c+id$ and comparing the terms. But no factors seem to cancel and the expression becomes complicated to analyze. I assume there should be a simpler, more direct method to analyze and come to a conclusion that $|w|<1$ is $|z|<1$.

Btw, this expression looks like $\arctan{z}-\arctan{a}$. Does that do anything?

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If you multiply $w$ by $\bar{w}$, you obtain $$|w|^2=\frac{|a|^2+|z|^2-2\mathrm{Re}(a\bar{z})}{1+|a|^{2}|z|^{2}-2\mathrm{Re}(a\bar{z})}.$$ The denominator is nonzero, since $\mathrm{Re}(a\bar{z})\leq|a||z|$, which means $1+|a|^{2}|z|^{2}-2\mathrm{Re}(a\bar{z})\geq 1+|a|^{2}|z|^{2}-2|a||z|=(1-|a||z|)^{2}>0$. Also, since $0<(1-|a|^{2})(1-|z|^{2})=1-|a|^{2}-|z|^{2}+|a|^{2}|z|^{2}$, we have that $|a|^{2}+|z|^{2}<1+|a|^{2}|z|^{2}$, so $|w|^2<1$.

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    $\begingroup$ Shouldn't the denominator be $ 1 + |a|^2 |b|^2 $? And isnt there a possibility of $ |a|^2 + |b|^2 < 2Re(a \bar z))? $ $\endgroup$ – Shashank Holla Dec 1 '16 at 13:43
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    $\begingroup$ Thanks for pointing out my error in the denominator (edited). There isn't a chance of this: $0\leq|a-z|^{2}=|a|^{2}+|z|^{2}-2\mathrm{Re}(a\bar{z})$. $\endgroup$ – RideTheWavelet Dec 1 '16 at 16:11
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We have to show that $$(z-a)(\bar z- \bar a)=|z-a|^2<|1-a \bar z|^2=(1-a \bar z)(1-\bar a z)$$ that is $$|z|^2+|a|^2-2\mbox{Re}(a\bar z)<1+|a|^2|z|^2-2\mbox{Re}(a\bar z)$$ and finally $$(1-|a|^2)(1-|z|^2)>0$$ which holds if $|a|<1$ and $|z|<1$ (or $|a|>1$ and $|z|>1$).

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Note that : $$z=i \rightarrow w=f(i)=\frac{i-\alpha}{1+i\alpha}=i $$ $$z=-i \rightarrow w=f(-i)=\frac{-i-\alpha}{1-i\alpha}=-i $$ $$z=1 \rightarrow w=f(i)=\frac{1-\alpha}{1-\alpha}=1$$ $$z=-1 \rightarrow w=f(i)=\frac{-1-\alpha}{1+\alpha}=-1$$ hence f maps |z|=1 to |w|=1 but $$ w=f(0)=-\alpha \rightarrow |w|=|\alpha|<1 $$ So f maps |z|<1 to |w|<1

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if you factorize denominator in ($\bar z$){1/($\bar z$) - $a$} and then as ($\bar z$){$z$ - $a$} , you will be able to see it now.

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  • $\begingroup$ Can you please explain? You did multiply and divide by $ \bar z $ in the denominator. Then? $\endgroup$ – Shashank Holla Dec 1 '16 at 13:40
  • $\begingroup$ now denominator is having one factor which is same as numerator. $\endgroup$ – Arpit Jain Dec 1 '16 at 13:42
  • $\begingroup$ so if you leave the point where $z$ = $a$, then equation reduces to $w$ = $1/$bar\z$$, that is $z$. $\endgroup$ – Arpit Jain Dec 1 '16 at 13:45
  • $\begingroup$ But it is given that |z|<1 not equal to 1. So $1/ \bar z $ is not equal to z $\endgroup$ – Shashank Holla Dec 1 '16 at 13:47
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    $\begingroup$ yes, also if $ |z| $ $<$ $1$ , $z$ < 1/$bar \ z$ $\endgroup$ – Arpit Jain Dec 1 '16 at 13:53

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