Z plus 1 times z plus 1, 1 times 1 is 1, and 1 plus 1 is 2.

So times z plus 1, times z plus 1.

(t 1) times (z 2 minus z 1) is going to look like this. This is (t minus 1) times (z 2 minus z 1).

1, 1, z, z looks like this z would look like, z would look like that.

(t time z 2 minus z 1). Is this equal to the argument of (z 2 minus z 1)?

This right here would be T times Z 2 minus Z 1. Now what would T minus 1 times Z 2 minus Z 1 look like?

This z plus 1 cancels out with one of these z plus 1's.

And, let's draw the vectored Z 2 minus Z 1 so let's say that this here is the vector and lets draw the vector Z 2 minus Z 1 now what would T times Z 2 minus Z 1 look like?

This is (t times z 2 minus z 1) and the green one, just to be clear is (z 2 minus z 1) is that.

Is it true that the argument of T times (z 2 minus z 1), is it true that that's the same thing as the argument of t minus 1 times (z 2 minus z 1).

This is z plus 1 squared.

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Because the derivative of z is 1.

The next z is equal to the z before it squared plus 1.

Well, the antiderivative of 1 with respect to z is just z, right?

Z 2 Z 1. So this is equivalent to saying that the argument of T times Z 2 minus Z 1, I'm just leveraging some of the algebra that we already did.

So this statement is equivalent to the statement that the argument of (z minus z 1) which is (t times z 2 minus z 1) leveraging the algebra from the last video.

This is (z 2 minus z 1) That's that angle. (t times z 2 minus z 1) is just going to be the scaled down version of it what we have in orange.

So the z intercept's going to be 1.

So this is equal to 7z times z plus 1, times z plus 2, times z minus 6.

So the claim is that this should be equal to the argument of (t minus 1) times (z 2 minus z 1).

Everything simplified to 7 times z minus 6 over 4 times z plus 1.

If y is 1 and z is 2, it's going to be 1 2.

If y is 0 and z is 1, it's going to be 0 ( 1).

But we could rewrite this thing right here as being equal to 1 minus T, I'm just factoring out the (1 minus t) times (z 1 minus z 2).

So, it's going to be a scaled down version of Z 2 minus Z 1.

So lets just say z is equal to 1.

Coefficients of the z terms are 1, 3, and 4.

As opposed to just 1 over z, y and x.

That choice B, the argument of z z 1 is equal to the argument of z z 2. Now lets think about this a little bit.

Look this is the same thing as abs( 15) abs( 46) abs( 29)

So we could rewrite this as 8 times z plus 2 times minus 1 times z minus 2.

We already did some work we simplified what (z z 1) is equal to. It's equal to T times

8 divided by 4 is 2 times so we can definitely factor out a z here, so we get z times z squared plus 2z plus 1.

So we're going to be scaling Z 2 minus Z 1 by a negative number. So this thing right over here.

Members elected for a three year term beginning on 1 January 1995 z

We have (z 2 minus z 1) its argument would be this angle right over here. It would be this angle five.

It's 5 times x minus 1 plus or maybe I should say minus 1, so it's plus minus 1 times y minus 2 plus minus 1 times z minus 3 is equal to 0.

So let's think a little bit about this right now. So, once again, we've already figured out what (z minus z 1) is equal to.

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. Choice D is telling us... so we don't know, we have to see if it's true that the Argument of (z minus z 1) is equal to the argument of (z 2 minus z 1)

So the argument in T time Z 2 minus z 1 is equal to, so I'm just rewriting statement B. Statement B is saying that this should be equal to the argument of... now what's (z minus z 2? z minus z 2 we've figured out was this thing right over here.

Actually, I have abs too.

It's not like that electronic abs machine where, 15 minutes a month, you get washboard abs.

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