Complex Number Proof Questions

Since $|alpha|=1implies alpha.overlinealpha=1$. Now, substitute $1$ as $alpha.overlinealpha$ in the numerator.

$$begin{align}left|frac{alphaoverlinebeta-1}{alpha-beta}right|&=|alpha|left|frac{overlinebeta-overlinealpha}{alpha-beta}right|\&=left|frac{-overline{(alpha-beta)}}{alpha-beta}right|\&=1end{align}$$

For question (14), $frac{|alphabeta'-1|}{|alpha-beta|}$=|$alpha $| $frac{|beta'-alpha'|}{|beta-alpha|}$(since, |$alpha $|=1),= $frac{|beta-alpha|'}{|beta-alpha|} $ =1.(as ${|beta-alpha|'} $ =|$beta-alpha$|) (I use prime notation instead of bar notation!)

Let $alpha,beta in mathbb{C}$ such that $alpha neq beta$ and $|alpha| = 1$. Consider the following expression:

$$left|frac{alpha beta'-1}{alpha-beta}right|$$

where I'm using prime notation for the complex conjugate since the overhead bar is annoying. Now, the correct way to do this is to recognize that for any $z in mathbb{C}$:

$$|z|^2 = z cdot z'$$

Now, we know that:

$$left|frac{alpha beta'-1}{alpha-beta}right| = frac{|alpha beta'-1|}{|alpha-beta|}$$

So, we can consider the numerator and denominator separately. Tackling the numerator first, we have:

$$|alpha beta'-1|^2 = (alpha beta'-1)(alpha beta'-1)' = (alpha beta'-1)((alpha beta')'-1) = (alpha beta'-1)(alpha'beta-1)$$

$$|alpha beta'-1|^2 = alpha alpha'beta beta'-alpha'beta-alpha beta'+1$$

$$ |alpha beta'-1|^2 = beta beta' -alpha' beta-alpha beta'+1 = |beta|^2-alpha'beta-alphabeta'+1$$

Now, let's work on the denominator:

$$|alpha-beta|^2 = (alpha-beta)(alpha-beta)' = (alpha-beta)(alpha'-beta') = 1-beta alpha' -beta' alpha + |beta|^2$$

Notice that:

$$|alpha-beta|^2 = |alphabeta'-1|^2$$

So, it follows that:

$$left|frac{alphabeta'-1}{alpha-beta}right| = 1$$

since the modulus of a complex number is greater than or equal to $0$. As for your other question, you should post another question regarding that. Also, I'd actually just recommend editing this so that Question 14 is within the actual post you've made. That'll make things a little more organized.

If there's anything you're confused about pertaining to Question 14 or what I have done above, let me know.