Submitted by stens on Sun, 05/07/2017 - 17:22
Forums:
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.