Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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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 can use DeMoivre's Theorem to calculate complex number roots. In many cases, these methods for calculating complex number roots can be useful, but for higher powers we should know the general four-step guide for calculating complex number roots. In order to use DeMoivre's Theorem to find complex number roots we should have an understanding of the trigonometric form of complex numbers.
Let's talk about how to find the roots of a complex number. We'll start with an example. Find the cube roots of 8i. I want to begin this by setting up an equation, z cubed equals 8i. Remember, the cube root of 8i would be a number that when cubed gives you 8i so all the cube roots have to satisfy this equation so I'm looking for solutions to this equation.
Now let's assume that the cube roots z are of the form r cosine theta plus i sine theta that is let's assume they're all in trig form. If they are, then I can just cube them using DeMoivre's Theorem and I also want to write 8i in trig form it's actually pretty easy because 8i the point is 8 units away from the origin so the modulus is 8 and its argument where we have a lot of choices but the most obvious choice is pi over 2. But let's remember that we could also add 2pi to that and that would also be a choice. We could add another 2pi and that will be a choice another 2pi and so on so let's keep that in mind but right now write cosine of pi over 2 plus i sine pi over 2. Alright, let's expand this let's simplify this using the DeMoivre's Theorem, we get r cubed cosine of, remember you multiply the argument by 3, cosine 3 theta plus i sine 3 theta and that equals 8cosine pi over 2 plus i sine pi over 2. Now for these two sides to be equal I need r cubed to equal 8 and I need 3 theta to equal pi over 2, now remember it doesn't just have to be pi over 2 it could be pi over 2 plus 2 pi or pi over 2 plus 4 pi, pi over 2 plus 2n pi any even multiple of pi, so first of all this r cubed equals 8 means r has to be 2 right, remember I'm looking for real number answer that's the that's going to be the length the modulus of my roots all of them will have a modulus of 2. What about the argument? I divide both sides by 3 and I get pi over 6 plus 2n pi over 3. When n equals 0 I'll get pi over 6 so one argument I could use is pi over 6. Let's start with that one, so one root would be z equals the modulus of 2 times cosine of pi over 6 plus i sine pi over 6. Cosine of pi over 6 is root 3 over 2 so this is 2 times root 3 over 2 oops plus and the sine of pi over 6 is a half so i times a half, 2 times root 3 over 2 is root 3, 2 times a half is 1 so I get i, sorry about that, and that's one of my roots root 3 plus i.
Now I get a second root if I let n equal 1. If n equals 1 then I'm adding 2pi over 3 to pi over 6, 2pi over 3 is the same as 4pi over 6 so 4 pi over 6 plus pi over 6 is 5pi over 6 that's my new argument so z equals 2 cosine 5pi over 6 plus i sine 5pi over 6 and I get 2 times the cosine of 5pi over 6 is minus root 3 over 2 and the sine of pi 5pi over 6 is a half again so I get minus root 3 plus i that's my second root.
My third root I get when n equals 2. When n equals 2 I have 4pi over 3 and adding 4pi over 3 which is the same as adding 8pi over 6, pi over 6 plus 8pi over 6 is 9 pi over 6 and 9pi over 6 is the same as 3pi over 2 so z equals 2 cosine 3pi over 2 plus i sine 3pi over 2 and this one's easy 3pi over 2 is this downward direction. The cosine of 3pi over 2 is 0 and the sine of 3pi over 2 is -1 so I get 2 times 0 plus i times -1 in other words -2i and it turns out that I'm done.
If I calculated for n equals 3 I'd end up getting the exact same root I got here. If I calculate for n equals 4 I'd get this one, n equals 5 I'd get this one, I keep cycling through these over and over again it turns out that they're only 3 distinct roots 3 distinct cube roots of any complex number. The number of roots equals the index of the roots so a fifth the number of fifth root would be 5 the number of seventh roots would be 7 so just keep that in mind when you're solving thse you'll only get 3 distinct cube roots of a number. And in addition to that let's take a look of the graph of these numbers I've plotted them out here, notice all 3 of them have a modulus of 2 they are at the same distance from 0 and they're all symmetrical they have 3 full rotation symmetry the angle between consecutive roots is 120 degrees this always happens with roots they always have this symmetry and they always have the same length. This is something to keep in mind when you're solving for the roots of the complex number.
Unit
Polar Coordinates and Complex Numbers