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!