Challenging complex numbers problem (2 of 3)

In the last video, we saw that choice A is true. Now let's see if B, C, and D are also true. So choice B, they're telling us that the argument, let me rewrite it over here-- I'll do it in another color, their choice B is telling us that the argument of z minus z1 is equal to the argument of z minus z2. Now let's think about this a little bit. We already did some work. We simplified what z minus z1 is equal to. It's equal to t times z2 minus z1. So this is equivalent to saying that the argument of t times z2 minus z1. I'm just leveraging some of the algebra that we already did, because this thing is the same thing as what's inside of the absolute value sign, which is the same thing as this thing over here. So the argument, t times z2 minus z1, is equal to-- so I'm just rewriting statement B. Statement B is just saying that this should be equal to the argument of-- now what's z minus z2? Z minus z2, we figured out was this thing over here. Actually, let's say it's this thing right over here. It's equal to this thing right over here. And we could rewrite this thing as being equal to-- I'm running out of real estate here-- but we can rewrite this thing right here as B equal to 1 minus t, I'm just factoring out the 1 minus t, times z1 one minus z2. Or so we can get in the same form here, let's multiply this times negative 1 and this times negative 1. So I'm multiplying by negative 1 twice. So I'm not changing the number. So this is going to be equal to, or the statement is claiming this is equal to, the argument of t minus 1-- let me write it a little bit neater than that. So the claim is that this should be equal to the argument of t minus 1 times z2 minus z1. So let's think about this. And just remember, all I did here is I multiplied this by negative 1, and this by negative 1. If I multiply two things times negative 1, it's equivalent to multiplying by negative 1 twice, which is just multiplying it by one. So I was able to swap both of these. So is this true? Is it true that the argument of t times z2 minus z1-- is it true that that's the same thing as the argument of t minus 1 times z2 minus z1? So let's just think about it a little bit. Let's draw an Argand diagram here. That's the imaginary axis. This is the real axis. And let's draw the vector z2 minus z1. So let's say that this right here is the vector. Let me draw it like this. This right here is the vector, z2 minus z1. Now what would t times z2 minus z1 look like? So t, we learned, is between 0 and 1. So it's going to be a scaled down version of z2 minus z1. So this right here-- so who knows what it is. This right here would be t times z2 minus z1. Now what would t minus 1 times z2 minus z1 look like? Well t minus 1, remember t is between 0 and 1, so t is less than 1. So t minus 1-- this right here is going to be a negative number. It's going to be a negative number. So we're going to be scaling it by a negative number. So we're going to be scaling z2 minus z1 by a negative number. So this thing right over here, t minus 1 times z2 minus z1 is going to look like this. This is t minus 1 times z2 minus z1. Now the arguments are just the angles between each of these numbers and the real axis. So the argument for this thing right over here is going to be this angle. It's going to be phi right over here. So I could call that phi. But what's the argument for this thing? It's going to be that plus pi, or plus 180 degrees. We have to go all the way around. So they are not the same angle. Whatever number this is right over here, this number is going to be that, plus or even minus pi. You could even go that way. But they're definitely not going to be the same angle. So we can cross out B choice B as an option. But now that we're in this mode, let's see if we can tackle choice D, because it looks very similar. And then I'll probably do choice C in the next video, because I don't want to spend too much time in each video. So what answer choice D telling us? So let me write it down here. Choice D is telling us-- so we don't know, we have to see if it's true, that the argument of z minus z1 is equal to the argument of z2 minus z1. So let's think a little bit about this right now. So once again, they tell us, we already figured out what z minus z1 is equal to. It's equal to this business over here. So this statement is equivalent to the statement that the argument of z minus z1, which is t times z2 minus z1, leveraging our algebra in the last video, t times z2 minus z1. Is this equal to the argument of z2 minus z1? Well, once again, let's draw our Argand diagram here. Actually, we could leverage the same Argand diagram. In green right here, we have z2 minus z1. It's argument would be this angle right over here. It would be this angle, phi. This is z2 minus z1. That's that angle. T times z2 minus z1 is just going to be the scaled down version of it, what we have an orange. Now clearly, this is just-- t is positive, so it's not pointing in another direction. This is just a scaled down version of this. This vector and this vector, or this complex number and this complex number, are going to point in the same direction, so their angles are the same. T-- let me do it-- this orange vector is this right over here, or that orange complex number is this right over here. This is t times z2 minus z1. And then the green one, just to be clear, z2 minus z1, is that. They're clearly in the same direction. They clearly have the same argument. So choice D another correct choice. Now I'm going to leave you there. In the next video, we're going to see if choice C works.