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# Challenging complex numbers problem (3 of 3)

Video transcript

So we're in the home stretch. Let's see if part C is true, and
I'm running out of real estate. So let me copy and
paste it actually. Actually, maybe I could do
it over here to the left. I had some doodles here. Let me clear those out. So let's see if we can do
it over here to the left. Part C. So we have
this determinant, and it's claiming
that it equals 0, so we have to see
if that's true. And so we can actually try
to multiply these out and see if anything interesting happens,
but we could even better leverage some of the
algebra that we've already done for part A. In part A, we
figured out that z minus z1 is equal to t times z2 minus z1. So this thing right
over here, this is the same thing as
t times z2 minus z1, and the reason why
that's interesting is I have a z1 minus
z1 here, so I'm starting to have similar things. And then this thing
over here-- so if I take the conjugate of each
of these, the conjugate of z minus the conjugate
of z1, this is going to be equal to t
times the conjugate of z2 minus the conjugate of z1. And I know that
because when I do this, I'm swapping all of the
signs on the imaginary parts of the complex numbers. So if you were to go through
this entire process just swapping the imaginary
parts of the complex number, the end result would have
it's imaginary parts swapped, and that's exactly
what we have over here. We have the imaginary
parts swapped over here. So if you want to work it out,
set this equal to A plus BI or set this equal to A1 plus
B1, you can work it out, but I think it's a pretty
intuitive idea that we're just making all of the-- we're
just swapping the signs on all of the imaginary parts of
each of these complex numbers. Now, with this said,
this determinant becomes pretty simple. This becomes t
times z2 minus z1. This becomes t times the
conjugate of z2 minus z1. This down here is z2 minus z1. And then this over here is
the conjugate of z2 minus z1. And so what is this determinant? It's going to be
this times this. So it's t times z2
minus z1 times z2 minus z1 times the
conjugate of those. So it's that times that
minus this times this, so minus t z2 z1, z2 minus
z1 times the conjugate of z2 minus the conjugate of z1. Now, this is exactly
equal to this. These things are obviously
going to cancel out, and we're clearly
going to get 0. So C is also true. So the correct answers
to this original problem were A, C, and D.