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

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?

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 $|w|^2<1$.

• Shouldn't the denominator be $1 + |a|^2 |b|^2$? And isnt there a possibility of $|a|^2 + |b|^2 < 2Re(a \bar z))?$ – Shashank Holla Dec 1 '16 at 13:43
• Thanks for pointing out my error in the denominator (edited). There isn't a chance of this: $0\leq|a-z|^{2}=|a|^{2}+|z|^{2}-2\mathrm{Re}(a\bar{z})$. – RideTheWavelet Dec 1 '16 at 16:11

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 $|z|>1$).

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

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.

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