Question

Question #693fa

Answer 1

`8sqrt(2)(cos((3pi)/4)+isin((3pi)/4))` or as `8sqrt(2)e^(((3pi)/4)i)`.

Explanation:

This doesn't use DeMoivre's exactly, but I'll answer the question.

Given `a+bi` we can find `r=sqrt(a^2+b^2)` and `hat theta=tan^-1(|b/a|)`.

In this case `r=sqrt((-8)^2+8^2)=8sqrt(2)`.
For `theta` we know the angle is in QII because of the ordered pair `(-8,8)`, and we know that `hat theta=tan^-1(1)=pi/4`. That means `theta=pi-pi/4=(3pi)/4`.

So we can write the complex number in polar form as `8sqrt(2)(cos((3pi)/4)+isin((3pi)/4))` or as `8sqrt(2)e^(((3pi)/4)i)`.