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Finding the Roots of a Complex Number - Problem 3 2,209 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.

I’m finding the roots of complex numbers and I have a particularly difficult example here. Find the square roots of z equals 7 plus 24l. It doesn’t sound that hard, but it’s going to turn out that you require some identities.

First of all, let’s put this in trigonometric form. So z equals 7 plus 24i. I want to find its modulus first. Its modulus is going to be the square root of 7² or 49, Plus 24² or 576. And that’s going to be the square root of 625 which is 25.

The argument, we know that cosine of the argument is going to be 7 over 25, and we know that the sine of the argument is going to be 24 over 25. And unfortunately, theta is not a special angle, so it’s not going to be as easy for us to solve for theta. So let’s leave it here. We will just write z as r which is 25 times cosine of theta plus i sine theta. We know that the square roots of z are going to have to satisfy the equation w² equals z. So w would be a square root of z.

We’ll need to have a trig form for w. Let’s make that w equals s times cosine of phi plus i sine of phi. The square of this by Demoivre's theorem, is going to be a² times cosine 2 phi plus i sine of 2 phi. And this has to equal z, which is 25 cosine theta plus i sine theta. And this means that s² is 25, and 2 phi is theta. Well theta plus 2n pi.

So because s, the modulus of w has to be positive, s has to be 5. And this means phi is going to be theta over 2 plus n pi. So let’s look at the square roots remember there going to be 2 distinct square roots of this complex number. One of them will have n equals 0.

When n equals 0, phi, the argument, is going to be theta over 2. So the square root will look like this, w equals the modulus 5 times cosine theta over 2 plus i sine of theta over 2. To get the cosine of theta over 2 or the sine of theta over 2, we would use the half angle identities. The cosine of theta over 2 is plus or minus the square root of 1 plus cosine theta over 2. And we have cosine theta back here.

Now let’s observe that theta is in the first quadrant. And so theta over 2 would be in the first quadrant and that means that the sine and cosine of theta over 2 are going to be positive. Let’s remember that because it helps us deal with this plus or minus, we know that we would get a plus here.

And again cosine theta is 7 over 25. Now if we simplify this, and you could simplify it by multiplying inside by multiplying inside by 25 over 25 and simplifying. You actually get 4/5. The sine of theta over 2 is plus or minus the square root of 1 minus cosine theta all over 2. Again cosine theta is 7 over 25, so I get 1 minus 7 over 25 over 2. And I choose plus because as I said before, the sine or theta is going to be positive. And it turns out that this simplifies to 3 over 5.

So this means that w is 5 times 4/5 plus i times 3/5. And this simplifies to 4 plus 3i. That is one of the square roots of 7 plus 24i. Let’s find the other one; n equals 1. For n equals 1, the argument is going to be theta over 2 plus pi. So w will equal 5 times the cosine of theta over 2, plus pi, plus i sine theta over 2 plus pi.

Instead of going through a really long process, of trying to figure out what these values are, let’s recall the add pi identities. They say that cosine of an angle plus pi is equal to minus cosine of that angle. And the same thing for sine. Sine of an angle plus pi is minus sine of that angle. Cosine of theta over 2, is 4/5 sine of theta over 2 is 3/5. So we get w is 5 times -4/5 plus i times -3/5. We get -4 and we get minus 3i. This is our second square root of 7 plus 24i.

Took a lot of work, required the sine cosine half angle identities, the add pi identities but here we are. These are the two square roots of 7 plus 24i.