Hi, I was just wondering if someone could put me in the right direction with rearranging this, from the LHS to the RHS equation? ieit = i eit – 1 1- e-it Thanks in advance!
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Originally Posted by Karty Hi, I was just wondering if someone could put me in the right direction with rearranging this, from the LHS to the RHS equation? ieit = i eit – 1 1- e-it That is not readable. Try again with simple text.
Do you mean $\displaystyle \frac{ie^{it}}{e^{it}-1}= \frac{i}{1- e^{-it}}$? If so, multiply both numerator and denominator by $\displaystyle e^{-it}$.
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