Current time:0:00Total duration:9:31
2003 AIME II problem 15 (part 3)
Where we left off in the last video, we were beginning to try to figure out what the 24 roots of 1 were. And then we were going to square those 24 roots, take the absolute value of their imaginary part, and then sum them up. Once again, let's just think about the roots that matter. So this over here, we figured out this is pi over 12. So let me-- this is pi over 12. That's the angle, I guess I should say. That's e to the pi over 12i. Let me just focus on the angle, here. This over here is 2 pi over 12, or pi over 6. This over here is 3 pi over 12, or pi over 4. This over here is 4 pi over 12, or pi over 3. And this right over here is 5 pi over 12. And then this, obviously, is 6 pi over 12, or pi over 2. We could keep going, but before we keep going-- and this will really just simplify the math a little bit-- remember, we're going to have to square these roots. So let's just think about if we have some complex number, a plus bi, let's think about what happens when we square it. This is going to be equal to a squared plus 2abi minus b squared. Or we could rewrite it as a squared minus b squared-- will be the real part-- and then, plus 2abi. 2ab is the imaginary part. And the reason why I even took the trouble of writing it this way is to realize that what we're going to do is we're going to square all of these complex numbers, and then take the absolute value of the imaginary part. So really, all that matters is the absolute value of 2ab, or really, 2 times the absolute value of ab. Now these guys all have analogs where either a or b would be negative. So if this guy is-- this right over here-- let's call this, a plus bi. If you had a minus bi, you would be down here. So if this guy is a plus bi, a minus bi is going to be over here. Or you could have negative a minus bi is going to be over here, or you could have negative a plus bi over here. But the reason why I did this is we showed you, just over here, all of these guys when you square them and you take the absolute value of the imaginary part of that squared value, it's all going to be the same. Because when you take-- it's going to be ab, absolute value is going to be ab. Negative ab absolute value just still going to be ab. So they're all going to be the same. So what we can do is-- and each of these guys are going to have the same analog. So what we could is just find the values for these guys one, two, three, four, for these five guys over here. And then just multiply whatever we get by 4, because they each have-- I guess there's four of these around the unit circle, and that'll save a lot of our work. Now, the other thing we want to think about-- we already said that we have to ignore 1, because we added that root. But even if we didn't remember to ignore it, it wouldn't matter, because 1 doesn't have an imaginary part. 1 squared is still 1-- does not have an imaginary part. We can also ignore an angle of pi over 2, or 90 degrees, because this has no real part. And you see over here, that when you square it, it's actually going to take you over here, and it won't have any imaginary part. When you square it it's 2 times the real times the imaginary part. This has no real part, so this is going to be 0. So this guy will also not contribute. So we really just care about these angles right over here. And then we'll square them, find the absolute value of their imaginary parts, and then we'll multiply everything by 4, because that'll correspond to these other-- if you took the imaginary or the real parts and made them negative. And it would take us all the way around the unit circle. So let's think about a little bit. Let me just write these over here. So the z's that we'll think about right now are going to be e to the pi over 12, e to the pi over 6, e to the pi over 4-- oh, e to the pi over 12i, I should say-- pi over 6i, pi over 4i, e to the pi over 3i, and then we have e to the 5 pi over 12i. Now we're going to square each of them, and it's nice to leave it in this exponential form when you square it. It's much easier to square these values. If you square this, you're just, essentially, multiplying the exponent times 2. So this is going to be e to the pi over 6i. This will be e to the pi over 3i-- we're just squaring them. So we're just taking the square of each of these values, each of these roots. So e to the pi over 3i, and then here you're going to have e to the pi over 2i. And over here, you'll have e to the 2 pi over 3i. And then finally over here, you'll have e to the 5 pi over 6i. Now these are the squared values of these roots here. Now, let's just think about their imaginary parts. So this guy right over here can be rewritten as cosine of pi over 6 plus i sine of pi over 6. So his imaginary part is sine of pi over 6. This guy's imaginary part-- if you expand it out, Euler's identity, Euler's formula. Cosine of pi over 3 plus i sine of pi over 3. So this imaginary part is just going to be sine of pi over 3. Here is going to be sine of pi over 2. Here is going to be sine of 2 pi over 3. And here is going to be a sine of 5 pi over 6. Now we just have to evaluate these guys, and take their absolute value, and then take the sum, and then multiply everything times 4. And we're essentially in the home stretch. So pi over 6 is 180-- if we were in degrees, and my brain has an easier time processing that. So, actually, let me draw another unit circle over here, just so we can visualize these angles. So now we have sine of pi over 6. Pi over 6 is the same thing as 30 degrees. So it looks like this. And we know that the sine of 30 degrees is 1/2. This is one, this is 1/2, this the cosine square root of 3 over 2. But this right over here, is 1/2. Sine of pi over 3. pi over 3 is the equivalent of 60 degrees. The sin over here is square root of 3 over 2. You can figure that out from 30, 60, 90 triangles. Then you have sine of pi over 2. Pi over 2 is that right over there. Well, the sine of that is just 1. So this is the imaginary part of this, or I guess, this would be-- essentially, this evaluates just to i, but the imaginary part, which is viewed as the coefficient on the i, which is kind of non-intuitive. You would kind of think it's the whole thing, but when people say the imaginary part. So this will just be 1. And then sine of 2 pi over 3. Let's see, that is-- so pi over 3 is 60 degrees. So you could view that is 120 degrees. So it's 120 degrees. So it's right over here. So it's going to have the same sin value as pi over 3, so it's going to be square root of 3 over 2. And then, sine of 5 pi over 6. Pi over 6 is 30 degrees. So this is sine of 150 degrees. So it's going to be just like that. And it's going to have the same sine value as pi over 6. So it's going to be 1/2, and lucky for us, these are all positive values. So let's just take the sum. And so we have 1/2 plus 1/2 is 1, plus another 1 over here is going to be two, plus the square root of 3 over 2 plus square root of 3 over 2 is square root of 3. Now remember, we just did it for this quadrant over here. We have to do it for all of the quadrants, so we just need to multiply everything times 4. So the sum of the absolute value of the imaginary part of the square of the roots is 8 plus 4 square roots of 3. And going back to the original problem, we got the answer here. This is 8 plus 4 square roots of 3. I want to make sure I didn't mess it up. Right. 8 plus 4 square roots of 3. So if we want find m plus n plus p, it is 8 plus 4 plus 3, which is equal to 15. And we're done.