Question

\[ \frac{e^{ik}-1}{1-e^{ik}}=\frac{1-e^{i2k}}{e^{i2k}-1} \]

Answer 1

This is obvious when you multiply by both denominators. However, which a in a/a do you multiply the LHS to get the RHS?

Answer 2

\[\frac{e^{ik}-1}{1-e^{ik}}\cdot\frac{e^{ik}+1}{1+e^{ik}}=\frac{e^{i2k}-1}{1-e^{i2k}}\cdot\frac{-1}{-1}=\frac{1-e^{i2k}}{e^{i2k}-1}\]

Answer 3

Goodness I can be a little bit special sometimes. Thanks!