How Do You Solve z^3 = sqrt(3) - i Using De Moivre's Theorem?

[Matty R]

Hello

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

Homework Statement

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

Homework 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

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.

 

[songoku]

Hi Matty

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

z³ = 2 cis (-π / 6)

r³e^i3θ = 2e^{-i(π / 6)}

Comparing coefficient :

1)
r³=2

2)
e^i3θ = 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

 

[Mark44]

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

 

[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.

Hi Matty

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

z³ = 2 cis (-π / 6)

r³e^i3θ = 2e^{-i(π / 6)}

Comparing coefficient :

1)
r³=2

2)
e^i3θ = 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

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:

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

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

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

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

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

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

 

[Mark44]

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

 

[Char. Limit]

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

 

[HallsofIvy]

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:

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

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

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

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

"$cis(\theta)$", 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 $e^{i\theta}$.

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