# Homework Help: De Moivre's Theorem : z^3

1. Feb 26, 2010

### Matty R

Hello

I'm really stuck on this question, and was hoping someone could help me.

1. The problem statement, all variables and given/known data

Find all complex numbers, z, that satisfy $$z^3 = \sqrt{3} - i$$

2. Relevant equations

$$z^n = r^n cis (n\theta)$$

$$r = \sqrt{a^2 + b^2}$$

$$\theta = tan^{-1}(\frac{b}{a})$$

Rotate anticlockwise by $$\frac{2\pi}{n}$$ radians

3. The attempt at a solution

$$z^3 = \sqrt{3} - i$$

$$r = \sqrt{(\sqrt3)^2 + (-1)^2} = 2$$

$$\theta = tan^{-1}(\frac{-1}{\sqrt3}) = -\frac{\pi}{6}$$

$$z^3 = 2cis(-\frac{\pi}{6})$$

This is where I go wonky.:shy:

$$n = \frac{1}{3}$$

$$(z^3)^{\frac{1}{3}} = 2^{\frac{1}{3}}cis(\frac{1}{3}.-\frac{\pi}{6})$$

$$z = \sqrt[3]{2}cis(-\frac{\pi}{18})$$

Now, from what I've seen in the lectures, I'm supposed to add $$\frac{2\pi}{\frac{1}{3}}$$ radians, then $$\frac{4\pi}{\frac{1}{3}}$$ radians, then $$\frac{6\pi}{\frac{1}{3}}$$ to get $$2\pi$$, and thus complete one revolution (360 degrees).

I've seen an Argand Diagram for this question divided into sections of 120 degrees, but I'm so confused.

Could anyone help me please? I'd very much appreciate it.

Thanks.

2. Feb 26, 2010

### songoku

Hi Matty

Ok, you have z3 = 2 cis (-π / 6), then using De Moivre's Theorem :

z3 = 2 cis (-π / 6)

r3ei3θ = 2e-i(π / 6)

Comparing coefficient :

1)
r3=2

2)
ei3θ = e-i(π / 6)
3θ = -(π / 6) ---> this is where you add 2π, 4π, ...or add 2kπ, where k is integer

so it will be :
3θ + 2kπ = -(π / 6) ---> You can also add 2kπ on RHS

Then find θ using several value of k

3. Feb 26, 2010

### Staff: Mentor

It's easier to work with an angle with positive measure, namely 11pi/6. You want values of z such that their 3rd powers are 2cis(11pi/6). If you have a complex number in polar coordinates, it's easy to calculate powers of that number by raising the modulus of the number to that power, and multiplying the angle (arg) by that power.

For example 1 + i = sqrt(2)cis pi/4, so (1 + i)2 = (sqrt(2))2cis pi2 = 2i.

In your problem, can you work backwards and find a complex number whose cube is sqrt(3) - i? All three of the numbers are spread equally around the circle 120 deg. apart

4. Feb 27, 2010

### Matty R

Thank you so much for the replies.

I think I may have worked it out now, but I'm still a bit confused.

Ok then.

$$z^3 = 2cis(-\frac{\pi}{6})$$

$$z^n = r^ne^{in\theta}$$

$$z^3 = r^3e^{i3\theta} = 2e^{i(-\frac{\pi}{6})}$$

$$3\theta = -\frac{\pi}{6} = (-\frac{\pi}{6}) + 2k\pi$$

$$\theta = \frac{(-\frac{\pi}{6}) + 2k\pi}{3} = -\frac{\pi}{18} + \frac{2k\pi}{3}$$

$$z = (\sqrt[3]{2})cis(-\frac{\pi}{18} + \frac{2k\pi}{3})$$

$$OR = (\sqrt[3]{2})cis(-\frac{\pi}{18})$$

$$OR = (\sqrt[3]{2})cis(-\frac{\pi}{18} + \frac{2\pi}{3})$$

$$OR = (\sqrt[3]{2})cis(-\frac{\pi}{18} + \frac{4\pi}{3})$$

Should they be written like this instead, seeing as I've used the exponential form earlier:

$$z = (\sqrt[3]{2})e^{i(-\frac{\pi}{18} + \frac{2k\pi}{3})}$$

$$OR = (\sqrt[3]{2})e^{i(-\frac{\pi}{18})}$$

$$OR = (\sqrt[3]{2})e^{i(-\frac{\pi}{18} + \frac{2\pi}{3})}$$

$$OR = (\sqrt[3]{2})e^{i(-\frac{\pi}{18} + \frac{4\pi}{3})}$$

Would that be $$(\sqrt3 - i)^{\frac{1}{3}}$$?

5. Feb 28, 2010

### Staff: Mentor

Well, yes, but that doesn't tell your what the three cube roots are.

6. Feb 28, 2010

### Char. Limit

I W-A'd it... and only got 1 answer.

7. Mar 1, 2010

### HallsofIvy

"$cis(\theta)$", which is short for $cos(\theta)+ i sin(\theta)= e^{i\theta}$ is primarily an engineering notation. Which is better to use depends on what kind of course this is. If it is a mathematics course being taught be a mathematician, I would say definitely use $e^{i\theta}$.