Challenging complex number problems
IIT JEE Complex Numbers (part 1) 2010 IIT JEE Paper 1 Problem 39 Complex Numbers (part 1)
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- Let z1 and z2 be two distinct complex numbers, and let z = (1-t) z1 + t z2, for some real number with
- t being between 0 and 1. If Arg (w) denotes the principal argument of a nonzero complex number w, then you know
- the argument is the angle of, kinda, the position vector specifiying the complex number in 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 our complex number, if that is z, its argument, the argument of z is going to be
- this angle right over here. So phi = Arg (z). That's all they're telling us in the second part.
- Now, let's go through each of these and see (inaudible) are true, and just to be clear
- this is one of those problems where more than one of these choices might be correct, so lets see if they are true.
- So let's first figure out what |z-z1| is. So |z-z1|, is going to be, so let me write. This is going to be, lets 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 - t z1 + t z2, 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 have, we can
- factor out a t over here, so this is going to be equal to, |t (z2 - z1)|. So that is |z - z1|, this first term over here.
- Let's figure out, |z - z2|. I'm going to color code it. |z - z2| is equal to, well z is just this thing up here
- let me just write it out, so it is (1-t) z1 + t z2, that's z. And from that we want to subtract z2. So
- |(1-t) z1 + t z2 - z2|. So, what can we do over here? We can take a t-1 out from over here, we can factor out
- the z2 out of these two terms. So this is going to be equal to the magnitude of (1-t) z1, thats this term right over here,
- plus, let me do it in another color, plus (t-1) 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, |t (z2 - 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?
- Well I have a 1-t and a t-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-t) z1, and so that I have a 1-t here, I'm just going to put a negative out front.
- So -(1-t), I just swap 'em, -(1-t) is the same thing as positive (t-1), z2.
- And then this thing over here, since I have the (1-t) out there, this becomes the magnitude of,
- I'm just factoring out the (1-t), |(1-t)(z1-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! So t is just a scalar, and t is between 0 and 1.
- They 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 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, actually let me not skip a step. This is going to be the same thing as |t| |z2-z1|,
- 'cause this is just scaling it, plus the |1-t||z1-z2|. Now let me be clear, |z1-z2| is going to be equal to |z2-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-z1| there.
- The reason why I'm doing it, is so I get the same thing here, and I can factor it out. So |z2-z1|, it's the same thing as |z1-z2|
- 'cause we're not, it's just going in the other direction. Now, what can we do here?
- Well, | t |, remember t is positive, so this is just going to be equal to t. |1-t|, once again that's positive, so that's
- just going to be 1-t. So we can factor, this business out, so we're going to get t + (1-t), times |z2-z1|.
- Now the t's cancel out here, we just got a 1 out front, so this is just equal to |z2-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.
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