Re: Complex Analysis - Cube roots of -8

First of all, you'll need a 2, somewhere.

|z| = cis(x/3) really doesn't make quite enough sense. It is as if you ALMOST know where you are going.

Start with

Just apply the theorem from DeMoivre.

for k = 0, 1, 2

k = 0 leads to

k = 1 leads to

k = 2 leads to

One for free. You show us the next one.

Re: Complex Analysis - Cube roots of -8

I think you're right; I ALMOST know where I am going! OK... I began with which is where I thought I was off track because of the zero in the real part. However, I should have known that was ok because -8 has nothing in the imaginary part... At any rate, moving on, and . But you say one for free and I should show you the next one. Aren't these three roots all of the roots of -8? Thank you so much for your help and your kind attitude.

Re: Complex Analysis - Cube roots of -8

That was one problem for free, not one cube root of -8.

How are you managing square roots and imaginary values?

Re: Complex Analysis - Cube roots of -8

I think you need to know that and .

Re: Complex Analysis - Cube roots of -8

Thank you; I thought that was the end of it! I managed the square roots with imaginary values with a graph of a right triangle and de Moivre's. This one threw me because it was just a line and there was no imaginary value, but you helped me see that it was basically the same as the other; I just had trouble with the angles. Thanks!