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.
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If you mean: is , the answer is no. Suppose . Then, by cancelling we get . By multiplying the LHS by , we get . Finally we end up with , which is a contradiction.
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... Either that, or you have a typo, or which leads to the result you've given
Last edited by Archie Meade; April 25th 2010 at 08:31 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
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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