00:03
Hello, we have to sketch the points in the complex plane satisfying this condition.
00:09
This condition is z plus 2 minus 4i mod which is equal to z minus minus 2 plus 4i.
00:36
This mod is less than equal to mod z minus 0.
00:45
This thing is the distance of z from the point minus 2 plus 4i and this thing is the distance of z from the point 0.
00:59
So it says that the distance of z from the point this this point will be less than equal to the distance of z from the origin.
01:08
So if you have the complex plane minus 2 plus 4i say this is this point minus 2 plus 4i let this be this point then let us join this point to the origin and draw the perpendicular bisector of this line segment.
01:50
Then the perpendicular bisector is the set of points which are equidistant which are equidistant from minus 2 plus 4i and the origin.
02:09
It says that the distance of z has to be equal or less than equal to the distance from the origin.
02:17
This is the set of points this half of the complex plane this line divides the complex plane into two halves.
02:27
This half of the complex plane including this line are the set of points which are this half of the complex plane including this bisector is the set of points whose distance from this point is less than equal to the distance from the origin.
03:02
So this is the solution set.
03:03
Now let us look at the second problem.
03:08
Here we also have to plot the set of points which satisfy this condition.
03:14
Now this condition z z bar plus we can write okay we can write i as minus i bar z it will be clear why we are writing it like this plus i we can write as minus of the conjugate of i minus i bar and then we can write this as minus i times z bar.
04:18
Now let us add i times i bar to this side and let us add now i times i bar i bar is minus i so i times minus i is plus 1...