Complex Polar Forms, Sin and Cos Angles

**tangibleLime** · September 21st 2011, 02:55 PM

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.

**alexmahone** · September 21st 2011, 02:59 PM

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 $\theta$ .

**tangibleLime** · September 21st 2011, 03:04 PM

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?

**alexmahone** · September 21st 2011, 03:08 PM

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 $[0,\ 2\pi]$ .

**Plato** · September 21st 2011, 03:11 PM

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 $2\exp \left( {\frac{{i\pi }}{3} + \frac{{2\pi ik}}{3}} \right),~~k=0,1,2$

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