Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $z,w \in \mathbb{C}$ s.t. $|z|,|w| < 1$. How would you show that $\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w|$?

I calculated that

\begin{align*} &\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w| \\ &\Rightarrow \frac{|z|^2-2Re(\bar{w}z)-|w|^2}{1-2Re(\bar{w}z) + |w|^2|z|^2} \leq (|z|+|w|)^2 \\ &\Rightarrow -2Re(\bar{w}z) \leq 2|z||w| + (|w|^2|z|^2 - 2 Re(\bar{w}z))(|z| +|w|)^2\ \end{align*}

If $2Re(\bar{w}z) < 0$, then $2|z||w| \geq 2Re(\bar{w}z)$ and $(|w|^2|z|^2 - 2 Re(\bar{w}z)) \geq 0$ imply that the inequality is true.

I haven't been able to figure out what happens in the case that $2Re(\bar{w}z) >0$ though.

On a separate note, whenever I come across these types of problems, I always try to multiply everything out and see if I can get enough terms to cancel so that the inequality becomes obvious.

I'm not quite sure if that's the smartest way to go about things, though and if there is some intuition that I should be using that I don't know about.

share|improve this question
1  
Aren't you using what you are trying to prove? I guess reverse the implications. –  PEV Feb 2 '11 at 5:33
1  
Well, I reduced what I am trying to prove using equivalences into an expression and scenario that I could justify. That is valid, right? –  user1736 Feb 2 '11 at 6:00
1  
Yes, but then you should be very careful to write $\iff$ (equivalence) at each step, instead of $\implies$. –  Dan Petersen Feb 2 '11 at 18:58
    
Fair point. Thanks for pointing that out! –  user1736 Feb 2 '11 at 23:36
add comment

2 Answers 2

up vote 2 down vote accepted

As Christian Blatter mentioned, this has a geometric interpretation in terms of hyperbolic metrics, but you can also do it directly: First note that $${|z - w|^2 \over |1 - \bar{w}z|^2} = {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2}$$ (You made a small error in your calculation). Next, observe that $|z|^2 + |w|^2 \leq 1 + |w|^2|z|^2$, since this is equivalent to $(1 - |w|^2)(1 - |z|^2) \geq 0$. Thus the above expression is of the form ${\displaystyle {A + c \over B + c}}$, where $0 \leq A \leq B$ and both $A + c$ and $B + c$ are nonnegative. Such an expression increases as $c$ increases. Since $- 2Re(\bar{w}z)$ is at most $2|w||z|$, we thus have $$ {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2} \leq {|z|^2 + 2|w||z| + |w|^2 \over 1 + 2|w||z| + |w|^2|z|^2}$$ $$ = {(|z| + |w|)^2 \over (|w||z| + 1)^2}$$ Showing this is at most $(|z| + |w|)^2$ is equivalent to showing $(|w||z| + 1)^2 \geq 1$, which is clearly true.

share|improve this answer
add comment

A hint: Think in terms of hyperbolic distance. If $d(z,w)$ is the hyperbolic distance between $z$ and $w\in D$ then $|z-w|/|1-\bar w z|= \tanh d(z,w)$, and in particular $|z|=\tanh d(0,z)$.

Another hint: You may assume $w=\rho>0$ and $z=r e^{i\alpha}$. Now compute $|z-w|^2$ and $|z-{1\over \bar w}|^2$ by means of the cosine theorem and show that the quotient of the two is largest when $t:=\cos\alpha=-1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.