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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.