Q: If $\alpha$ and $\beta$ are different complex numbers with $ |\beta|=1$, then find $ \bigl|\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\bigr|$

[correct answer is $1$]

My proof: Let $ \alpha = a+ \iota b$ and $\ \beta = c + \iota d$

Since $|\beta| = 1 \implies \beta = 1,-1$ [opening modulus]

Required: $ \cfrac{|\beta - \alpha|}{|1- \bar \alpha \beta|}$

Case 1: When $ \beta =1$
$$ \implies \cfrac{|(1) - \alpha|}{|1- \bar \alpha (1)|} $$ $$ \implies \cfrac{|(1) - (a+\iota b)|}{|1- (a-\iota b)|} = \cfrac{|(1) - a -\iota b|}{|1- a+ \iota b|} = \cfrac{|(1 - a) -(\iota b)|}{|(1- a) + (\iota b)|} $$ Since modulus of a complex number of the form $z = x+ \iota y$ is $|z| = \sqrt{x^2+y^2} $
$$\implies \require{cancel} \frac{\cancel{\sqrt{(1-a)^2+(-b)^2}}}{\cancel{\sqrt{(1-a)^2+(-b)^2}}} = 1$$

Case 2: When $\beta = -1$
$$ \implies \cfrac{|(-1) - \alpha|}{|1- \bar \alpha (-1)|} $$ $$ \implies \cfrac{|(-1) - (a+ \iota b)|}{|1 + (a - \iota b)|} = \cfrac{|(-1 - a) - (\iota b)|}{|(1 + a) - (\iota b)|} $$ [ the below part is where I think that my solution is wrong ] $$ \implies \cfrac{\sqrt{(-1)(1+a)^2+(-b)^2}}{\sqrt{(a-1)^2+(b)^2}} $$ $$ \implies \cfrac{ \sqrt{ 1+a^2-2a+b^2 } }{ \sqrt{ 1+a^2-2a+b^2 } } $$

If my solution is wrong, how can it be corrected?

one error I identified when opening the identities in numerator and denominator:
Numerator: $ \bigl((-1)^2 + a^2 -2(-1)(a) \bigr) +b^2 = 1+a^2+2a+b^2$
Denominator: $ \bigl( (a)^2 + (-1)^2 + 2(-1)(a) \bigr) + b^2 = 1+a^2 -2a +b^2 $

I don't understand why this way can't work.

  • 2
    $\begingroup$ Why does $|\beta| = 1$ impply that $\beta = 1$ or $-1$? There can be many choices for $\beta$, for example $i,-i$, etc. Note we are speaking of the complex modulus, not the real modulus, since $\beta$ is a complex, not necessarily real number. $\endgroup$ – астон вілла олоф мэллбэрг Aug 9 '18 at 16:38
  • 1
    $\begingroup$ $$ z=\cfrac{\beta - \alpha}{1- \bar \alpha \beta} $$ $$ z \bar z= \left (\cfrac{\beta - \alpha}{1- \bar \alpha \beta} \right ) \left (\cfrac{\bar \beta - \bar \alpha}{1- \alpha \bar \beta} \right ) =\cfrac{\beta \bar \beta-\bar \alpha \beta -\alpha \bar \beta+\alpha \bar \alpha} {1-\bar \alpha \beta -\alpha \bar \beta+\alpha \bar \alpha \beta \bar \beta} $$ $$ z \bar z =\cfrac{|\beta|^2-\bar \alpha \beta -\alpha \bar \beta+|\alpha|^2} {1-\bar \alpha \beta -\alpha \bar \beta+|\alpha|^2|\beta|^2} $$ So if $|\beta|=1$, unless $\alpha=\beta$, $|z|=1$ $\endgroup$ – NMK Aug 9 '18 at 17:06

So, you have a couple of things wrong here

First, $\beta \in \mathbb{C}$, so we can't simply assume $\beta \in \{1, -1\}$, what about $\beta \in \{ z \in \mathbb{C}: \lvert z\rvert=1\}$ which contains all of the complex numbers of the form $z = 1*[cos\theta+i*sin\theta]$

As far as the proof, it is quite simple. You should look up "modulus properties" and "complex number properties"

Things I used:

$\lvert z \rvert = \lvert \bar{z} \rvert$

$\lvert z \rvert \lvert w \rvert = \lvert zw \rvert$

$z*\bar{z} = \lvert z \rvert ^2$

Let me get you started.

We know $\lvert \beta \rvert = 1$, so we can also know that $\lvert \bar{\beta} \rvert = 1$ by property 1.

We use this to take:

$ \dfrac{\lvert\beta-\alpha\rvert}{\lvert1-\bar{\alpha}\beta\rvert}*1 = \dfrac{\lvert\beta-\alpha\rvert}{\lvert1-\bar{\alpha}\beta\rvert}*\lvert\bar{\beta}\rvert $

Then we use property 2 to get $\lvert \bar{\beta} \rvert$ into our numerator.


Finally, it is up to you to find why this is $1$


In $\mathbb{C}$ if $|\beta|=1$ then $\beta$ can be anywhere on the unit circle centred at $0$. In particular,

$$\vert{\beta}\vert=1 \iff \beta\in \{e^{it}:t\in[0,2\pi)\}$$

A general trick when working with $\vert z \vert$ is to use the fact that $\vert z \vert^2=z \bar{z}$. So:

$$\begin{align} \left\lvert \cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right\rvert^2 &= \left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right) \overline{\left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right)} \\ &= \left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right) \left(\cfrac{\bar{\beta} - \bar{\alpha}}{1- \alpha \bar{\beta}}\right) \end{align}$$

By expanding this bracket and cancelling, you should find your solution!


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