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IIT JEE complex root probability (part 2)

Video transcript
In the last video, we started on this problem right here, where we said let omega be a complex cube root of unity, and omega cannot be equal to 1. And so we figured out what all of the complex cube roots of unity were. We figured out-- well, we knew that 1 was one of them, and we used that to factor out this third degree equation right over here. And we figured out the other roots, cube roots, of unity were negative 1 plus or minus the square root of 3i over 2. And we said, look, the problem said let omega be one of the, or be a complex cube root of unity. So I just picked it to be the negative version. So I said let omega be this thing right over here. Now, just to kind of explore the space a little bit, we said, OK, what's omega squared? And we figured out that it was actually the conjugate of this. It's the plus version of this. So we got omega squared right over here. And then I kind of wasted your time a little bit because we know what omega cubed is. We know that omega is a cube root of unity, so omega cubed must be unity. But it wasn't harmful, I guess, to go through the process. It shows you sometimes my brain gets into a rut and just does stuff that it doesn't have to do. But we figured out-- we multiplied this times omega to show that, hey, it is definitely equal to 1. So what we were able to do is set up a situation, and we will use this to think about the next part of the problem. So omega is equal to negative 1/2. Or maybe I should say omega to the first power is equal to negative 1/2 minus the square root of 3 over 2i. Omega to the second power-- let me write it over here-- omega to the second power is equal to negative 1 plus the square root of 3 over 2i. And then omega to the third power, and that was pretty straightforward, is equal to 1. And I touched on this in the last video. What is omega to the fourth power? Omega to the fourth power is going to be omega to the third times omega. So it's going to be omega again. So this is also going to be-- let me scroll to the left a little bit-- this is also going to be omega to the fourth power. Omega to the fourth power is the same thing as a omega to the first power, which is this. Now, what's to the fifth power? Well, it's going to be omega times the fourth power, so it's omega times this. Well, omega times this is the same thing as omega squared, so it's going to be that. So this is omega to the fifth power. And what's omega to the sixth power? Well, that's just the same thing as taking omega cubed and squaring it, so this is also going to be 1. So this is omega to the sixth. Or you could just use, it's this times omega because it's to the fifth power, and we saw that that's also equal to 1. And the reason why I went up to the sixth power is because, if you'll remember the old problem, we're rolling a die. And on that die-- I'm assuming it's a normal six-sided die-- we're going to get values between 1 and 6. And we're going to take omega to the different powers, and we're going to see if it equals to 0. And we want to find the probability of this being equal to 0. So let me write this down. We want to find the probability that omega-- and we're just using the omega that we picked because they said it's one of the complex roots that isn't 1-- so we want to find the probability that omega to the first die of the roll plus omega to the second die of the roll plus omega to the third die of the roll, that when you take their sum, that that is equal to 0. This is what we need to figure out. So to figure this out, let's just figure out what combinations of powers of omega will even add up to 0. How can we even add them up to 0? I mean, they all have to cancel out some way. And if you look at them, it looks pretty interesting of how they might cancel out. If I take one version of this, the only way that they really can cancel out is if I take one version of this, and add it to this, and then add it to this. Let me show you. If I take negative 1/2 minus the square root of 3 over 2i, and I add it to this, and I add it to this, negative 1/2 plus the square root of 3 over 2i, and then I add it to 1, and then I add it to this, and then I add it to this, what do I get? Well, you're going to have this guy and this guy are going to cancel out. Negative 1/2 plus negative 1/2 are going to be equal to negative 1. You add negative 1 to 1, it's going to be equal to 0. So essentially, what it was asking us, what's the probability that I get, for each of these terms, that I get one each of each of these powers of omega? Or another way of thinking about this, what's the probability-- so if we think of it this way. Let's think of it this way. r1 could be equal to-- we're going to get omega here, which is the same thing as omega the fourth power. So r1 could be 1 or 4. And this is in the situation where the first one gives us omega. The second one is omega squared, or the second one gives us this value. And then the third one gives us 1. So r2 would be 5 or 2. And then r3 would be equal to 3 or 6. Now, I want to be very clear. We could swap these around. There's actually six ways that you could-- instead of this being the first roll, this roll could be 1 or 4, and then this one could be 5 or 2, and then this one could be 3 or 6. So there's actually six ways of doing this, I guess you could say. You could permute these six different ways, or 3 times 2 times 1. But we'll think about that in a second. But assuming that we want this way, where this first one is going to evaluate to this, and the second one is going to evaluate to this, and the third one is going to evaluate to this, what's the probability of that happening? And then we're going to multiply that 6 by 6 because there are six ways to arrange these terms right over here. So let's do that. So what's the probability of that happening? Well, the probability that r1 is a 1 or a 4, well, that's two values out of six, so the probability is 1/3 there. The probability that r2 is a 5 or 2, well, that's also going to be 1/3. There's two values out of a possible of six. The probability that r3 is a 3 or a 6, only two possibilities there. So two out of six possible faces of the die, so times 1/3. So the probability that this first one is going to evaluate to this value right over here, the second one is going to evaluate to this value, and the third one is going to evaluate to 1, is going to be 1/3 times 1/3 times 1/3, which is 1/27. And I touched on this already, there's six ways. You could rearrange this six ways. You have three terms, and you're putting them in three places. So in the first place, you could put three of the terms. In the second place, you have two terms left that you could put. And in the last place, you only have one term left. So there's 3 times 2 times 1 ways to arrange these things. We can arrange this 3 times 2 times 1 different ways, so there's six different arrangements. The probability of each is 1/27. So the probability under question, this thing over here, there's six arrangements. There's six arrangements of getting these things to be added up in this way, and the probability of each of them is 1/27. So it's equal to 6/27, and if you divide the numerator and the denominator by 3, it's equal to 2/9. And we're done. It wasn't too bad, I think. The probability of getting omega to the r1 plus omega to the r2 plus omega to the r3, where r1 and r2 and r3 are numbers obtained from rolling a fair die, the probability-- when you add all of these, it's going to be equal to 0-- is 2/9. And I thought that was a pretty neat problem.