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# DeMoivre's Theorem - Problem 2

###### 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 that asks; if z equals 1 plus i, express z to the -1, z to the 0, z², z³, and z to the 4th in rectangular form. We're going to use Demoivre's Theorem to do this, but first we have to get z into trig form, because Demoivre's Theorem only works with complex numbers in trig form.

We first have to find the modulus of z. And the modulus is going to be r equals, the square root of 1² plus the coefficient of i, 1² root 2. Then we have to find the argument. Now 1 plus i, this is actually pretty easy, it's going to be right here. The argument is going to be pi over 4. 1 plus pi is right on line y equals x right on the diagonal. So theta is pi over 4, and that means that z is root 2 times cosine pi over 4 plus i sine pi over 4.

Now once you know the modulus, and the argument, raising the power just means, taking the modulus and raising it to the power, and multiplying the argument by the power. So for example, let's start with z². Z² is going to be r², times cosine of 2 theta plus i sine 2 theta. Now here, r is root 2. Root 2² is 2, and theta, the argument is pi over 4. 2 times pi over 4 is pi over 2. Now, the cosine of pi over 2 is 0. So this is 2 times 0 plus i, and the sine of pi over 2 is 1. So this is just 2i in rectangular.

How about z³? Z³ will be r³. Root 2 is the modulus, cubed, cosine of pi over 4, the argument, times 3. So 3 pi over 4 plus i sine 3 pi over 4. Root 2 to the third power is 2 root 2. The cosine of 3 pi over 4 is negative root 2 over 2, and the sine of 3 pi over 4 is positive root 2 over 2.

So let's multiply through, we get the 2's cancelling minus root 2 times root 2 minus 2 plus i, and the 2's cancel again, and we get root 2 times root 2 which is 2. I can actually write 2i, and that's the rectangular form of z to the third.

Now z to the fourth. Root 2 to the 4th power times cosine of pi over 4 times 4 is pi. This one is going to be easy, plus i sine pi. Now root 2 to the 4th is the same as 2², 4. Cosine of pi is -1, sine of pi is 0. So this is just -4. So 1 plus i to the 4th power is -4.

Now what's really interesting about Demoivre's Theorem, is that also works, and not just for positive integer powers, it works for the 0th power, and the -1 power, any negative power, any integer power.

So let's try z to the 0. Now you might suspect z to the 0, just like real numbers, raised to the 0 are going to end up to 1, but let's see if that works out with complex numbers. So we've got root 2 to the 0, that's the modulus raised to the 0, and then we've got cosine of pi over 4 times 0. Cosine of 0, plus i sine 0. So it's 1 times cosine 0 which is 1, plus i times sine of 0 is 0. So it's just 1 times 1, which is 1. So z to the 0 is 1 as you might expect, and hope it to be. Z to the -1.

I think it's really interesting that this works for negative integer powers too. But you're going to get root 2 to the -1, and cosine of pi over 4, the original argument, times -1. So negative pi over 4 plus i sine negative pi over 4.

Now 1 over root 2 is the same as root 2 over 2. This is going to be cosine of negative pi over 4. Cosine is even, so that's the same as cosine of positive pi over 4, which is root 2 over 2. i times the sine of negative pi over 4 is the opposite of the sine of pi over 4, so minus root 2 over 2. Root 2 over 2, times root 2 over 2 is 2 over 4, one-half. Root 2 over 2, well it's the same thing, -1/2 this time though. So it's minus 1/2i. That's z to the -1. It's the reciprocal of 1 plus i. So now you can also take reciprocals using Demoivre's Theorem. Just make n, -1.

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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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