# Thread: Complex Polar Forms, Sin and Cos Angles

1. ## Complex Polar Forms, Sin and Cos Angles

Problem: Find all solutions of $z^{3} = -8$.

I can do the entire problem except for one part. After putting it into the correct form,

$|z|^{3}(cos3\theta+isin3\theta)$,

I do not know how to find the values of $cos3\theta$ or $sin3\theta$. I know that $|z|^3$ is 8, but I can't figure out the values of cos and sin.

Any help is appreciated.

2. ## Re: Complex Polar Forms, Sin and Cos Angles

Originally Posted by tangibleLime
Problem: Find all solutions of $z^{3} = -8$.

I can do the entire problem except for one part. After putting it into the correct form,

$|z|^{3}(cos3\theta+isin3\theta)$,

I do not know how to find the values of $cos3\theta$ or $sin3\theta$. I know that $|z|^3$ is 8, but I can't figure out the values of cos and sin.

Any help is appreciated.
$|z| = 2$.
So, $cos\ 3\theta+isin\ 3\theta=-1$.
Equate real and imaginary terms to find $\theta$.

3. ## Re: Complex Polar Forms, Sin and Cos Angles

Thanks,

So since $|z|^3 = 8$, I need $cos3\theta+isin3\theta$ to equal -1 to satisfy the initial equation where $z^3 = -8$. The only way to get -1 from $cos3\theta+isin3\theta$ is to have $cos3\theta = -1$ and $sin3\theta = 0$ to get rid of the imaginary number and return (-1 + i0).

Correct?

4. ## Re: Complex Polar Forms, Sin and Cos Angles

Originally Posted by tangibleLime
Thanks,

So since $|z|^3 = 8$, I need $cos3\theta+isin3\theta$ to equal -1 to satisfy the initial equation where $z^3 = -8$. The only way to get -1 from $cos3\theta+isin3\theta$ is to have $cos3\theta = -1$ and $sin3\theta = 0$ to get rid of the imaginary number and return (-1 + i0).

Correct?
Yes. You should get three values of $\theta$ in the interval $[0,\ 2\pi]$.

5. ## Re: Complex Polar Forms, Sin and Cos Angles

Originally Posted by tangibleLime
Problem: Find all solutions of $z^{3} = -8$.
Here is some notation: $\exp(i\theta)=\cos{\theta)+i\sin(\theta)$

So we can write $-8=8\exp(\pi i)$
The cube roots of that is then $2\exp \left( {\frac{{i\pi }}{3} + \frac{{2\pi ik}}{3}} \right),~~k=0,1,2$