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Challenging complex numbers problem (1 of 3)

Paper 1 Problem 39 from the challenging 2010 IIT JEE exam, about complex numbers. Created by Sal Khan.

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Video transcript

Let z1 and z2 be two distinct complex numbers. And let z equal, and they say it's "1 minus t times z1 plus t times z2, for some real number with t being between 0 and 1." And they say, "If the argument w denotes the principal argument of a nonzero complex number w, then?" And we know the argument is the angle of kind of the position vector specifying the complex number of the complex plane, and the real axis. Let me just draw that. So if we do an Argand diagram, this is the imaginary axis. And this is the real axis. If this is our complex number, if that is z, it's argument-- the argument of z-- is going to be this angle right over here. So phi is equal to the argument of z. That's all they're telling us in this second part. Now, let's go through each of these and see are they true. And just to be clear, this is one of those problems where more than one of these choices might be correct. So let's see if they're true. So let's first figure out what the magnitude of z minus z1 is. So the magnitude of z minus z1 is going to be the magnitude of-- so let me write-- this is going to be the magnitude of-- let's do the z part first. Let me write in the same colors. So if I would just focus on the z first, z is this thing up here. And let me distribute the z1. So it's z1 minus tz1 minus tz1, and then plus tz2. And from that, we want to subtract z1. So from that, we want to subtract z1, so minus Z1. And that and that cancel out. And let's see. We can factor out a t over here. So this is going to be equal to the magnitude of-- let's factor out a t times z2 minus z1. So that is the magnitude of z minus z1, this first term over here. Let's figure out the magnitude of z minus z2. I'm going to color code it. z minus z2 is equal to the magnitude-- well, z is just this thing up here. Let me just write it out. So it is 1 minus t times z1 plus t times z2, that's z. And from that, we want to subtract z2, so minus z2. So what can we do over here? We can take a t minus 1 out from here. We can factor out the z2 out of these two terms. So this is going to be equal to the magnitude of 1 minus t times z1, 1 minus t times z1. That's this term right over here. Plus-- let me do it in another color-- plus t minus 1, times z2. The magnitude of this part over here is this thing over there. Now, let's see what happens when we add this thing to this thing. That's what choice A is doing. We're adding this thing to this thing. Let's see if we can simplify it. So this becomes the absolute value, or the magnitude, of t times z2 minus z1 plus-- I'll do the whole thing here in magenta-- plus this thing over here. And this thing over here-- actually, before I even write it out-- how can we simplify that? We'll have a 1 minus t and a t minus 1. Well, I'll just write it out here. But I'm going to change it up a little bit. So this is equal to the magnitude of 1 minus t times z1. And so that I have a 1 minus t here, I'm just going to put a negative out front. So minus 1, minus t, I just swapped them. Negative 1 minus t is the same thing as positive t minus 1, z2. And then this thing over here, since I have the 1 minus t out there, this becomes the magnitude of, I'm just factoring out the 1 minus t, 1 minus t times z1 minus z2. So we have this thing. Let me copy and paste it. So, copy and paste. We have this thing plus this thing. This is what this expression A has simplified to. Now let's see if we can simplify that even more. Let's see if we can simplify it even more. So t is just a scalar. And t is between 0 and 1. They have told us that over here, t is between 0 and 1. So this is positive. This right here is a positive value. And then this right here is also going to be a positive value. t is greater than 0, and it is less than 1. So this is also going to be a positive value. And these are just scalars. So this is going to be the same thing. These are just scaling the magnitude. So this is going to be the same thing. These are positive values. So this is going to be t times the absolute value-- actually let me not skip a step. This is going to be the same thing as the absolute value of t times the absolute value of z2 minus z1-- because this is just scaling it-- plus the absolute value of 1 minus t, times the absolute value of z1 minus z2. Now let me be clear, the absolute value of z1 minus z2 is going to be equal to the absolute value of z2 minus z1. These vectors are just pointing in different directions, or these complex numbers. One is just the negative of the other. But their absolute values, or their magnitudes, are going to be the same. So let me write z2 minus z1, the absolute value of z2 minus z1, there. And the reason why I'm doing it is so I get the same thing here. And I can factor it out. So the absolute value of z1 minus z1 is the same thing as the absolute value of z1 minus z2, because it's just going in the other direction. Now, what can we do here? Well, the absolute value of t, remember t is positive, so this is just going to be equal to t. The absolute value of 1 minus t, once again, that's positive, so that's just going to be 1 minus t. So we can factor this business out. So we're going to get t plus 1 minus t, times the absolute value of z2 minus z1. Now, the t's cancel out here. We just got a one out front. So this is just equal to the absolute value of z2 minus z1. So we see that choice A does work. This plus this does indeed equal that. I'm going to leave you there. And in the next video, we're going to try out some of these other possibilities to see if they might also be true.