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Finding the Roots of a Complex Number - Problem 1 2,895 views

Teacher/Instructor 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.

Finding the roots of complex numbers here’s an example. Find the fourth roots of -16 and -16 is a complex number even though it is also real. Remember to find the fourth root we would set up an equation like this. Z to the fourth equals -16.

So, I’m going to imagine that my roots are complex numbers of the form; z equals r times the quantity cosine theta plus i sine theta. I can raise z to the fourth power using Demoivre's theorem. I would get r cosine theta plus i sine theta to the fourth power. And that’s going to be r to the fourth cosine 4 theta plus i sine 4 theta. Now on the right side I’ve got -16 which I also want to write in trig form. -16 has a modulus of 16, that’s its distance from the origin, and it's got an argument of pi. So 16 cosine pi plus i sine pi.

Now pi is not the only angle that will work, it's just one of the angles that works. You could also add 2 pi to that or 4 pi or 6 pi etcetera. So 16 cosine pi plus i sine pi. Let’s observe that, in order for these two sides to be equal, we need r to the fourth to equal 16. And we need 4 theta to equal pi. Actually 4 theta could equal pi plus any even multiple of pi.

So let’s divide both sides by 4 we get theta equals pi over 4, plus 2n pi over 4. And that’s pi over 4 plus n pi over 2. Let’s start looking at the roots of -16. First for n equals 0. When n equals 0, we get pi over 4 plus 0, so our argument will be pi over 4. And then we’ll get r equals 2, that’s going to be our modulus. 2 cosine pi over 4 plus i sine pi over 4. Cosine pi over 4 is root 2 over 2 and so is the sine of pi over 4. So this is 2 times root 2 over 2 plus i root 2 over 2. 2’s cancel and you get root 2 plus i times root 2.

That’s the first fourth root of -16. The second comes from n equals 1. When n equals 1, I’m adding pi over 2 to pi over 4. And I get 3 pi over 4 that’s going to be my argument. So z equals 2 cosine 3 pi over 4 plus i sine 3 pi over 4.

The cosine of 3 pi over 4 is negative root 2 over 2. Sine 3 pi over 2 is still 3 pi over 4 is still root 2 over 2. So this is 2 times negative root 2 over 2 ,plus I times root 2 over 2. Again the 2’s cancel and I get negative root 2 plus i root 2, so that’s the second. the 4th root of -16.

Let’s find the third, n equals 2. When n equals 2, I’m adding pi to pi over 4. So my new argument is going to be 5 pi over 4. I get z equals 2 times the cosine of 5 pi root 4 plus i sine 5 pi over 4.

Cosine of 5 pi over 4, negative root 2 over 2. So is the sine of 5 pi over 4. So it's 2 times negative root 2 over 2, plus i times negative root 2 over 2. It simplifies to negative root 2 minus i times root 2.

And finally for n equals 3, we are adding 3 pi over 2 which is the same as 6 pi over 4. Add that to pi over 4 you get 7 pi over 4. That’s our argument. Z equals 2, cosine 7 pi over 4 plus i sine 7 pi over 4. Cosine 7 pi over 4, is root 2 over 2. Sine of 7 pi over 4 is negative root 2 over 2. I get root 2 minus i times root 2.

This is my fourth and last fourth root of -16. Remember when you're taking fourth roots you're only going to expect four distinct roots. And we have them 1, 2, 3, 4. And here they are, I’ve graphed them on a complex plane. Here is -16 way out here, modulus of 16. Each of these little guys is going to have a modulus of just 2, the fourth root of 16.

And notice also that these guys have 4 fold rotational symmetry. They are all 90 degrees apart from one another. The roots of a complex number always have this kind of symmetry. So it’s one way to check to see that your answers are right plot them.

So once again these are the four, the fourth roots of -16.