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# DeMoivre's Theorem - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We're talking about powers of complex numbers. I have a problem here. If z equals 1/2 plus root 3 over 2i, express z², z³, z to the 4th, z to the 5th, and z to the 6th in trigonometric form.

Now to do this, I'm first going to have to convert z into trig form, and I'll use Demoivre's Theorem to come up with all these powers. It won't be real, so I'll just first convert z to trig form. To do that, I need to get the modulus of z.

Modulus r is going to be the square root of 1/2 squared plus (root 3 over 2)². That's 1/4 plus 3/4. The square root of 1/4 plus 3/4 is the square root of 1, which is 1. So this is a modulus of 1.

Now what's the argument? I'll call the argument theta. You know that the cosine of theta is the real value, 1/2 divided by the modulus. So it's 1/2, and the sine of theta is the imaginary coefficient root 3 over 2, divided by the modulus 1. So it's root 3 over 2. So what angle has a cosine of a half, and a sine of root 3 over 2? It's pi over 3, and that means, that z is 1 times cosine pi over 3 plus i sine pi over 3. So this is the number that will be raising to powers.

Let's start with z². Now we use Demoivre's Theorem; it's the modulus squared times cosine of 2 theta plus i sine 2 theta, where theta is pi over 3. Now the modulus is 1, so 1² is 1, and we'll get cosine of 2 times pi over 3, 2 pi over 3, plus i sine 2 pi over 3. Z³, again the modulus is 1, 1³ is 1,. So it's going to be cosine of (and I multiply the argument by 3) 3 times pi over 3 is pi. As you can see it's really easy.

Z to the 4th, it would be much harder to do this in rectangular form. Z to the 4th is 1 to the 4th which is 1, times cosine of 4 times pi over 3 plus i sine 4 pi over 3. Z to the 5th, 1 to the 5th which is 1, cosine of 5 pi to the 3, plus i sine 5 pi over 3.

Finally Z to the sixth is 1 to the sixth, which is 1, cosine of 6 times pi over 3 is 2 pi, and we have cosine 2 pi plus i sine 2 pi. So these are the second, third, 4th, 5th, and sixth powers of z. Now what I want to do is plot these on the complex plane along with z. All of these, because the modulus for all of them is 1, all of these are going to fit somewhere on the unit circle.

The first one has an argument of pi over 3, so it's about here, that's z. Z² had an argument of 2 pi over 3, so it was about here. Z³ had an argument of pi, so it ends up here. Z to the 4th, argument of 4 pi over 3, so down here. Z to the 5th and z to the sixth.

It's sort of interesting that when you take consecutive powers of a complex number, you always advance by the same angle. So in this case we keep advancing by pi over 3. That always happens, and so it gives the powers symmetric look.

Now you don't always have that all the powers are the same modulus. Very often the modulus will increase as you take powers, or decrease depending on the original modulus. But here the modulus was 1, so 1² is 1, 1³ is 1. So they all have the same distance from 0. These are the first six powers of z.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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