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# Challenging complex numbers problem (2 of 3)

Video transcript

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.