Question

Need help with De Moivre's Theorem!

1. Find (√3/2 - 1/2i)^4. Leave answer in rectangular form.

2. Find (√3 + i)^1/3.

3. Find 4√16i. Leave answer in rectangular form.

4. Find (-√3/2 - 1/2i)^3

Thank you!

Answer 1

\[\Large\rm \left(\frac{\sqrt3}{2}-\frac{1}{2}\mathcal i\right)^{4}\]Match these up with your trig functions:\[\Large\rm \cos \theta=\frac{\sqrt3}{2},\qquad\qquad \sin \theta=-\frac{1}{2}\]Cosine positive, sine negative,
that puts us in the 4th quadrant.

Do you know what angle these values correspond to?

Answer 2

Unfortunately, yes. I'm not good with this theorem. Everything else I am great at, but I'm horrid at this one.

Answer 3

@zepdrix
For cosine, 270 degrees?
For sine, I'm not sure.

Answer 4

No no. These values correspond to one angle.
Together, they put us as a specific location on the unit circle.

Answer 5

If that's too hard, you can think of it terms of tangent.\[\Large\rm \tan \theta=\frac{\left(-1/2\right)}{(\sqrt3/2)}=-\frac{\sqrt3}{3}\]Do you remember your tangent angles on the unit circle?

Answer 6

210 degrees?

Answer 7

Degrees? :(
Comeon bud! Join the big boys! It's all about the radians XD

Answer 8

7pi/6?

Answer 9

210 is in the third quadrant, that can't be right.
We need to be in the 4th quadrant. hmm

Answer 10

No it's going to be the `last` pi/6 before we start to repeat values.

Answer 11

Oh 11pi/6?

Answer 12

Ok good.

Answer 13

It would be easier to use the principle value between -pi and pi.
It'll just work out easier with our calculations I think.

So unwind one full rotation.
It brings us to -pi/6, yes?

Answer 14

Yes

Answer 15

So let's write our values in terms of our angle:\[\Large\rm \left(\color{orangered}{\frac{\sqrt3}{2}-\frac{1}{2}\mathcal i}\right)^{4}\quad=\quad \left(\color{orangered}{\cos \left[\frac{-\pi}{6}\right]+\mathcal i \sin\left[\frac{-\pi}{6}\right]}\right)^{4}\]

Answer 16

Then from here, we can finally apply De Moivre's Theorem.

Answer 17

Okay. Is r next to find?

Answer 18

No r for this problem.

Answer 19

r=1

Answer 20

Oh, ok

Answer 21

So, De Moivre's let's us bring the exponent inside, yes?
\[\Large\rm =\left(\cos\left[-\frac{4\pi}{6}\right]+\mathcal i \sin\left[-\frac{4\pi}{6}\right]\right)\]Simplifying,\[\Large\rm =\left(\cos\left[-\frac{2\pi}{3}\right]+\mathcal i \sin\left[-\frac{2\pi}{3}\right]\right)\]

Answer 22

And once again we have a special angle, so we can change these trig functions up.

Answer 23

Okay, that makes sense

Answer 24

So can we move onto the 2nd problem?

Answer 25

I don't think my teacher cares about special angles.

Answer 26

\[\Large\rm \left(\sqrt3+\mathcal i\right)^{1/3}\]So what I usually do is, I try to turn the stuff inside the brackets into something that will produce our special angle.

You can let me know if this is too confusing,
there is another approach we can take if it is.

So I start by factoring a 2 out of each term,\[\Large\rm 2^{1/3}\left(\frac{\sqrt3}{2}+\frac{1}{2}\mathcal i\right)^{1/3}\]From here, we have nice values that relate to one of our special angles.\[\Large\rm \cos \theta=\sqrt3/2,\qquad\qquad\qquad\sin\theta=1/2\]

Answer 27

In the examples we did in class, my teacher didn't mention special angles.

Answer 28

That doesn't make any sense.
You can't apply De Moivre's Theorem until you've written it in terms of Sine and Cosine....

Answer 29

For example, when finding (1 + i√3)^5 she got 32(cos(5pi/3) + isin(5pi/3))

Answer 30

I can show you how you would get that answer if you want.
You're missing the important step in which you find your special angel though.

Answer 31

Factoring out a 2,\[\Large\rm \left(1+\mathcal i \sqrt3\right)^5=2^5\left(\frac{1}{2}+\frac{\sqrt3}{2}\mathcal i\right)^5\]1/2 and sqrt3/2 is in the first quadrant,
it's ummmm, let's see cosine is the short one in this case, so it's pi/3\[\Large\rm 2^5\left(\cos\frac{\pi}{3}+\mathcal i \sin \frac{\pi}{3}\right)^5\]Then finally you can apply De Moivre's Theorem,\[\Large\rm 2^5\left(\cos\frac{5\pi}{3}+\mathcal i \sin \frac{5\pi}{3}\right)\]

Answer 32

I'll do it your way

Answer 33

Let's try it the other way I guess.
Maybe it will make more sense to you.\[\Large\rm \left(\sqrt3+\mathcal i\right)^{1/3}\]
Use this to find your angle theta,\[\Large\rm \tan \theta=\frac{imaginary}{real}\qquad\to\qquad \tan \theta=\frac{1}{\sqrt3}\]
Use this to find your radial value,\[\Large\rm r=\sqrt{(\sqrt3)^2+(1)^2}\]

Answer 34

Then you'll be able to write it in the form:\[\Large\rm \left(\sqrt3+\mathcal i\right)^{1/3}=r^{1/3}(\cos \theta+\mathcal i \sin \theta)^{1/3}\]

Answer 35

Ohhh. I see