Just wondering if it was possible for someone to show me how you get from this equation:
ie^(it)/e^(-it)-1
to this
i/1-e^(-it)
Cheers in advance! Its much appreciated.
If you mean: is $\displaystyle \frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}$, the answer is no.
Suppose $\displaystyle \frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}$. Then, by cancelling $\displaystyle i$ we get $\displaystyle \frac{e^{it}}{e^{-it}-1}=\frac{1}{1-e^{-it}}$. By multiplying the LHS by $\displaystyle \frac{-1}{-1}$, we get $\displaystyle \frac{-e^{it}}{1-e^{-it}}=\frac{1}{1-e^{-it}}$. Finally we end up with $\displaystyle -e^{it}=1$, which is a contradiction.
hi Karty,
you are losing a factor...
$\displaystyle \frac{ie^{it}}{e^{-it}-1}=\frac{-ie^{it}}{-(e^{-it}-1)}$
$\displaystyle =\frac{-ie^{it}}{1-e^{-it}}=-e^{it}\ \frac{i}{1-e^{-it}}$
Either that, or you have a typo,
$\displaystyle \frac{ie^{-it}}{e^{-it}-1}=\frac{ie^{-it}e^{it}}{e^{it}\left(e^{-it}-1\right)}=\frac{i}{1-e^{it}}$
or
$\displaystyle \frac{ie^{it}}{e^{it}-1}=\frac{ie^{it}e^{-it}}{e^{-it}\left(e^{it}-1\right)}$
which leads to the result you've given