Many students are stuck when they see something like \frac{1}{1-e^{\frac{i\theta}{3}}} for example. They are unsure what tto do and some of them attempt to rationalise it.

Here’s a little tip to resolving all such questions. 🙂

Given 1-e^{n\theta} for n can be anything, fractional or integer, we simply factorise e^{\frac{n\theta}{2}} from the expression. Notice that 1=e^{i0} so using simple indices, this gives

e^{\frac{n\theta}{2}}(e^{\frac{-n\theta}{2}}-e^{\frac{n\theta}{2}})

Our objective of doing this is to have (e^{\frac{-n\theta}{2}}-e^{\frac{n\theta}{2}}). Now notice that this is a very familiar form, its z^{*}-z form which gives us -2iy=-2isin{\theta} for this case! Isn’t that convenient!

e^{\frac{n\theta}{2}}(e^{\frac{-n\theta}{2}}-e^{\frac{n\theta}{2}})

=(cos{\frac{n\theta}{2}}+isin{\frac{n\theta}{2}})(-2isin{\frac{n\theta}{2}})

=-2isin{\frac{n\theta}{2}}cos{\frac{n\theta}{2}}-2isin{\frac{n\theta}{2}}isin{\frac{n\theta}{2}}

=-isin{n\theta}+2sin{\frac{n\theta}{2}}sin{\frac{n\theta}{2}}

=-isin{n\theta}+2sin^{2}{\frac{n\theta}{2}}

After some manipulation, we find the following result. In the event, have a fraction like one above, we still start with the same method. Hope this helps!

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