Question

Write the complex number `(-1-i)^6` on the `re^θ = r(cos θ + i sin θ)` form (polar form) and the `a+ ib` form?

This is the solution, but can someone explain how this happen?

Answer 1

We know `z=-1-i=re^(itheta)`

`r=sqrt((-1)^2+(-1)^2)=sqrt(2)`
`theta=tan^-1(1)=pi/4`

However, as `-1-i` is in quadrant 3, we must add `pi` to `theta` to get `(5pi)/4`

We now have `z=sqrt2e^(i(5pi)/4)`

`(-1-i)^6=z^6=(sqrt2e^(i(5pi)/4))^6=(sqrt2)^6(e^(i(5pi)/4))^6=8e^(i(15pi)/2)=8e^(i(3pi)/2)`

In terms of rotation `(15pi)/2-=(3pi)/2`

`8e^(i(3pi)/2)=8(cos((3pi)/2)+isin((3pi)/2))=8(0-i)=-8i`