00:01
We're being asked to solve the given equation for z.
00:02
So we can treat this just like we would a regular equation.
00:05
So the first thing i'm going to do is remove my preemphasies.
00:08
So for the first on the left hand side, we'll distribute the z.
00:12
Well, 4 times z is 4z, and 2i times z is positive 2iz.
00:18
Now, because there's only a positive sign in front of the second set of parentheses, we can just bring both of those terms down.
00:24
So plus 7 minus 2i.
00:26
Now on the right hand side, we'll distribute the z again.
00:29
4 times z is 4 z and negative i times z is negative i z and in front of that last of the preemphasy we just have a plus sign so we can bring down both of those terms plus three plus five i so the next thing i need to do is because we're solving for z is to get my z term on the left hand side so to do this i'm going to add i z to its like term well two i z plus i z is three i z so this is going to leave us with 4z plus 3i z plus 7 minus 2i, which is equal to 4 z plus 3 plus 5i.
01:07
So now i'm going to move all the terms that don't have a z to the right hand side.
01:11
So i'll begin by subtracting 4z from both sides, but that's going to cancel out on both sides.
01:16
So that's great.
01:18
Then i'll subtract the constant 7 from both sides, so those will cancel.
01:22
And i'll add 2i to both sides.
01:25
So now on the left -hand side of our equation, we just have 3i -z...