Cool trick to solve questions in polar form

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!

Leave a Comment

16 + nineteen =

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search