formula rearranged

**Karty** · April 25th 2010, 07:01 AM

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.

**Tikoloshe** · April 25th 2010, 08:05 AM

If you mean: is $\frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}$ , the answer is no.

Suppose $\frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}$ . Then, by cancelling $i$ we get $\frac{e^{it}}{e^{-it}-1}=\frac{1}{1-e^{-it}}$ . By multiplying the LHS by $\frac{-1}{-1}$ , we get $\frac{-e^{it}}{1-e^{-it}}=\frac{1}{1-e^{-it}}$ . Finally we end up with $-e^{it}=1$ , which is a contradiction.

**Archie Meade** · April 25th 2010, 08:18 AM

Originally Posted by Karty

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.

hi Karty,

you are losing a factor...

$\frac{ie^{it}}{e^{-it}-1}=\frac{-ie^{it}}{-(e^{-it}-1)}$

$=\frac{-ie^{it}}{1-e^{-it}}=-e^{it}\ \frac{i}{1-e^{-it}}$

Either that, or you have a typo,

$\frac{ie^{-it}}{e^{-it}-1}=\frac{ie^{-it}e^{it}}{e^{it}\left(e^{-it}-1\right)}=\frac{i}{1-e^{it}}$

or

$\frac{ie^{it}}{e^{it}-1}=\frac{ie^{it}e^{-it}}{e^{-it}\left(e^{it}-1\right)}$

which leads to the result you've given

**Wilmer** · April 25th 2010, 08:18 AM

Originally Posted by Karty

ie^(it)/e^(-it)-1

As you show it:
ie^(it)/e^(-it) - 1
= ie^(it)e^(it) - 1
= ie^(2it) - 1

**Karty** · April 25th 2010, 09:04 AM

my mistake, there was an error in the formula:

ie^(-it)/e^(-it)-1

to this

i/1-e^(-it)

sorry to confuse, and thanks for taking the time to answer my qs!

Karty

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