 ###### 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.

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# More Roots of Complex Numbers - 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.

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We're talking about finding the roots of complex numbers. Here's a way a problem can be worked. They might be a little confusing. Solve z to the fifth minus 32i equals 0. Now this is really just the same as solving the equation z to the 5th equals 32 i. What we're really doing is finding the fifth roots of 32 i. That's what the solutions of this equation are.

Don't be put off by solving an equation, you're still finding the roots of a complex number. The first thing we need to do is put 32i in trig form. If you plot it it's really easy to see what the trig form would be. You got a modulus or distance from 0 of 32. You've got an argument of Pi root 2. This is going to be 32 cosine Pi root 2 plus i sine Pi root 2.

Remember how to find the roots quickly. Let's start with the primary root. Call it z1. First of all I'm looking for 5th roots. I take the modulus and raise it to the 1/5 power. 32 to the 1/5 power is 2. The modulus of my roots are going to all be 2. Then I take the argument of the original number, Pi over 2, and I divide it by 5 or multiply 1/5. I get Pi over 10. This is 2 cosine Pi over 10 plus i sine Pi over 10. That's the first or primary root.

The rest of the root, successive roots are going to be found by taking this argument and adding 2 Pi over 5. I've got to add 2 Pi over 5 which is the same as 4 Pi over 10 to this number, to get the next root. Z2 is going to be 2, same modulus cosine of Pi over 10 plus 4 Pi over 10 is 5 Pi over 10. 5 Pi over 10 is Pi over 2.

This one can be simplified. Cosine of Pi over 2 is 0. Sine of Pi over 2 is 1. This becomes 2i. I'm just going to simplify them when it's easy to do so. Then z3, the next one, I add 4 Pi over 10 to Pi over 2. Now Pi over 2 remember is 5 Pi over 10. I get 9 Pi over 10. Cosine of 9 Pi over 10 plus i sine 9 Pi over 10.

The fourth; 2 cosine, again add 4 Pi over 10, I get 13 Pi over 10, plus i sine 13 Pi over 10.

The fifth one, the fifth and last, remember there are only going to be 5 distinct fifth roots. 2 cosine of; I add 4 Pi over 10 and I get 17 Pi over 10, plus i sine 17 Pi over 10. That's it. Each of these numbers is a fifth root of 32i and therefore a solution of this equation.