# Prove that $\left|\frac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1$ [duplicate]

Question: Prove that $\left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1$ if $|z_1|\lt1$, $|z_2|\lt 1$

My solution:

$$$$\left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1 \\ \implies \left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|^2\lt 1 \\ \implies \left( \dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right)\overline{\left({\dfrac{z_1-z_2}{1-z_1\bar{z_2}}}\right)} \lt 1 \\ \implies |z_1|^2+|z_2|^2 \lt 1 +|z_1|^2|z_2|^2 \\ \implies (|z_1|^2-1)(1-|z_2|^2) \lt 0$$$$

Now as $|z_1| \lt 1$, so $(|z_1|^2-1) \lt 0$ and similarly $(1-|z_2|^2) \gt 0$, hence we can safely conclude that the above inequality holds.

Now the book that I am solving from also gave the same solution it just further factored $(|z_1|^2-1)(1-|z_2|^2)$. This doesn't seem a good enough proof for me. If anyone can suggest a proof which doesn't simply verify the statement to be proved.

P.S.:- This question has also been asked here, but I don't think I have the same query as the OP of that post.

## marked as duplicate by Guy Fsone, Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 23 '17 at 0:34

• all of your steps are two-sided implications, that's why it works ok – gt6989b Jul 10 '16 at 12:25
• So, are you saying that this is the only method. – user350331 Jul 10 '16 at 12:26
• no, i am saying you don't really need another one, this is good enough – gt6989b Jul 10 '16 at 12:28
• Of course not. I am actually thinking about a very geometric proof in terms of distances: note that after clearing the denominator you get that $|z_1-z_2|<|1-z_1\overline{z_2}|$ and the first is the length of the segment between the two points, whereas the second is the distance between the point $1$ and that product, which can be better understood if you switch to polar coordinates... maybe using some sort of triangle inequality you might find another proof. – b00n heT Jul 10 '16 at 12:31
• If you don't like your equivalence chain then calculate $| 1-z_1\bar{z_2} |^2 - |z_1-z_2|^2 = ... = (1 - |z_1|^2)(1 - |z_2|^2)$. – Martin R Jul 10 '16 at 12:59

For $\left|z\right|\le1$ and $-1\le x\le1$, we have \begin{align} 1-\left|\frac{(1+z)x}{1+zx^2}\right|^2 &=1-\frac{(1+z)x}{1+zx^2}\frac{(1+\bar z)x}{1+\bar zx^2}\\ &=\frac{\left(1-x^2\right)\left(1-\left|z\right|^2x^2\right)}{1+2\mathrm{Re}(z)x^2+\left|z\right|^2x^4}\\ &\ge\frac{\left(1-x^2\right)\left(1-\left|z\right|^2x^2\right)}{\left(1+\left|z\right|x^2\right)^2}\\[6pt] &\ge0\tag{1} \end{align} Assume, wlog, that $\left|z_1\right|\le\left|z_2\right|$. Then, setting $z=-\frac{z_1}{z_2}$ and $x=\left|z_2\right|$ in $(1)$, we get \begin{align} \left|\frac{z_1-z_2}{1-z_1\bar{z}_2}\right| &=\left|\frac{\left(1-\frac{z_1}{z_2}\right)\left|z_2\right|}{1-\frac{z_1}{z_2}\left|z_2\right|^2}\right|\\[6pt] &\le1\tag{2} \end{align}

As has been pointed out, this only solves the problem when we maximise the numerator, which of course doesn't give the full answer. At the moment I cannot see how to solve it in general, but I'll leave this answer just as a start.

Maybe one can try by fixing $r_1,\phi_1,r_2$ and seeing the difference as a function of $\phi_2$ and try to minimize it. But it's not what I was looking for

Note that two points $z_1=r_1e^{i\phi_1}$ and $z_2=r_2e^{i\phi_2}$ have the maximal distance if $\phi_2=\phi_1+i\pi$, i.e. they are symmetric with respect to the center of the circle. Their distance is then simply given by $$|z_1-z_2|=r_1+r_2.$$ but now $$z_1\overline{z_2}=r_1r_2e^{i\pi}=-r_1r_2$$ so that $$|1-z_1\overline{z_2}|=1+r_1r_2.$$ Now the proof follows by showing that for $r_i<1$ $$1+r_1r_2>r_1+r_2$$ Which is equivalent and follows from the fact that $$(1-r_1)(1-r_2)>0.$$

• Why have you considered the only condition for their maximal distance, how would one prove if they weren't diametrically opposite. – user350331 Jul 10 '16 at 12:56
• This is not right. Maximizing the numerator does not maximize the fraction. In fact here it turns out that maximizing the numerator also maximizes the denominator, and it's simply not true that $A/B\le (\max A)/(\max B)$. – David C. Ullrich Jul 10 '16 at 13:09
• Both of you are right, and at the moment I cannot generalize this to a full proof. – b00n heT Jul 10 '16 at 13:12
• I'll be happy to see it – b00n heT Jul 10 '16 at 14:42
• I was writing it then i found it to have many loopholes which I dont think I would be able to stand up to say that they were right – user350331 Jul 10 '16 at 14:53

All denominators in your calculation are strictly positive, and multiplying an inequality with a positive number gives an equivalent inequality. Therefore, as already pointed out in the comments, the implications are two-sided and you can simply replace $\Longrightarrow$ by $\Longleftrightarrow$ to get a fully valid proof.

That is a common procedure if you try to prove something: Start with the claim and try to find a "simpler" equivalent statement. If that task was successful then you can try to find a "direct" proof.

In your case, a possible approach would be to calculate $$\lvert 1-z_1\bar{z_2} \rvert^2 - \lvert z_1-z_2 \rvert^2 = ... = (1 - \lvert z_1 \rvert^2)(1 - \lvert z_2 \rvert^2)$$ (which is valid for all complex number $z_1, z_2$). For $\lvert z_1 \rvert < 1, \lvert z_2 \rvert < 1$ the expression is positive, and $$\left\lvert \frac{z_1-z_2}{1-z_1\bar{z_2}}\right \rvert\lt 1$$ follows. As a "bonus", you see immediately what happens if both $z_1, z_2$ are outside of the unit circle, or one is inside and the other is outside, without repeating the calculation.