# 10 S3 Q6

The first part of the problem asks us to prove that
$\lvert e^{i\beta}-e^{i\alpha}\rvert = 2\sin\frac12(\beta-\alpha),$
but as far as I can see, we have the identity
$2\sin\frac12(\beta-\alpha) = \frac{e^{i\beta}-e^{i\alpha}}{ie^{i\frac12(\alpha+\beta)}},$
and the denominator of the RHS is generally not equal to $\pm 1$. Are we supposed to prove that the magnitude of the complex number $e^{i\beta}-e^{i\alpha}$ is equal to the RHS in the first equation instead?

### This is not 2010 S3 Q6. Do

This is not 2010 S3 Q6. Do you mean 2009 S3 Q6?

I'm not really sure what you're asking; the question is asking you to show that $|e^{i\beta} - e^{i\alpha}| = 2\sin \frac{1}{2}(\beta - \alpha)$, the (very standard) notation $|.|$ is the absolute value or magnitude of the complex number $e^{i\beta} - e^{i\alpha}$. What do you mean by "instead"?

### Yes you are right, it's from

Yes you are right, it's from the 2009 STEP. I guess I interpreted the absolute value as something like "up to a plus or minus sign", but that makes no sense when we are working with complex numbers (because there is no reason to believe that $e^{i\beta}-e^{i\alpha}$ is real, right?). With that, I believe the conclusion follows immediately from the identity above, because the absolute value of the denominator is $1$?

### Yep! (although I'm pretty

Yep! (although I'm pretty sure you can't just quote the identity, you'd have to derive it.)

Recall that $|z|$ is the distance of a number from the origin. In the case of $\mathbb{R}$ this being a real line we can only have $|z| = 1$ means $z=\pm 1$ as we are on a line if $z \in \mathbb{R}$. In $\mathbb{C}$ however, we are on a plane and $|z| = 1$ is a circle.

Underground Mathematics: Selected worked STEP questions

STEP Question database

University of Cambridge Mathematics Faculty: What do we look for?

University of Cambridge Mathematics Faculty: Information about STEP