Complex number question help

Question

How would I find the solutions in polar form to $$z^3=-8$$ Can you provide a step by step solution?

You go first. What was your first step? – Matthew Leingang Dec 30 '16 at 15:45 — Matthew Leingang, Dec 30 '16 at 15:45

score 2 · Answer 1 · answered Dec 30 '16 at 15:47

2

Notice that

$$-8=8e^{(2k+1)\pi i}$$

for $k\in\mathbb Z$.

Thus,

$$z=2e^{\frac{2k+1}3\pi i}$$

answered Dec 30 '16 at 15:47

score 1 · Answer 2 · answered Dec 30 '16 at 16:01

1

Without the polar form:

Notice that

$$z^3+8=(z+2)(z^2-2z+4).$$

Then the roots are $-2$ and $1\pm i\sqrt3$.

answered Dec 30 '16 at 16:01

score 0 · Answer 3 · answered Dec 30 '16 at 16:10

For the equation $Z^n=z_0$ we have n roots which are:

$Z_k = n\sqrt{r} (\cos \frac{\theta +2k\pi}{n}+i\sin\frac{\theta +2k\pi}{n})$ where r is modulus of $z_0$.

So, In your equation, you will get three values of $Z$ which will be $-2$ and $1\pm i\sqrt3$.

3 Answers3