Thread: Complex Polar Forms, Sin and Cos Angles

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

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

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

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

Any help is appreciated.

Originally Posted by tangibleLime

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

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

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

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

Any help is appreciated.

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

Thanks,

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

Correct?

Originally Posted by tangibleLime

Thanks,

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

Correct?

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

Originally Posted by tangibleLime

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

Here is some notation: $\displaystyle \exp(i\theta)=\cos{\theta)+i\sin(\theta)$

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