Complex Number Proof Questions

Question

$Q.14)$ If $\alpha,\beta$ are complex numbers where $\alpha\ne \beta$ and $|\alpha| = 1$. Prove that $\displaystyle\left|\frac{\alpha\overline\beta-1}{\alpha-\beta}\right|=1$. Image of question.

For $Q.14$ I tried separating the expression into real and imaginary components and then solving it but it does not lead to anywhere. What approach should I take to solve this question? Thanks.

Answer 1

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:

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

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

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

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:

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

So, it follows that:

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

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.

Answer 2

For question (14), $\frac{|\alpha\beta'-1|}{|\alpha-\beta|}$=|$\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!)

Answer 3

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

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