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Math Algebra all content Complex numbers Challenging complex number problems. Challenging complex numbers problem 1 of 3.

Challenging complex numbers problem fr of 3. Challenging complex numbers problem 3 of 3. Challenging complex numbers problem: Current time: 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.

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So let me copy and paste it girlfrlend. Actually, maybe I could do it over here to the left. I had some doodles.

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Let me clear those. So let's see if we can do it over here to the left. Part C.

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So we have this determinant, and it's claiming that cennai sex 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 dat that we've already done for part A.

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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.

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And then this thing over here-- so if 3 3 3 3 girlfriend for a day 3 3 3 3 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.

We have the imaginary parts swapped over. So if sex big dick black 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.

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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.

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And so what is this determinant? It's going to be this times. So it's t times z2 minus z1 times z2 minus z1 times the conjugate of.

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.

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. Up Next.