WEBVTT
00:00:01.280 --> 00:00:10.840
In this video, weβll learn how to express regions in a complex plane in terms of inequalities and how to interpret inequalities as regions in the complex plane.
00:00:11.680 --> 00:00:18.000
Itβs likely you all have worked extensively with regions in the real Cartesian plane, considering them as inequalities.
00:00:18.310 --> 00:00:20.570
And this video looks to extend those concepts.
00:00:21.250 --> 00:00:31.670
Weβll begin by considering regions defined by half lines, circles, and perpendicular line bisectors, before looking at how set notation can help us to find composite regions.
00:00:33.080 --> 00:00:38.030
Remember, there are two equations that are used to define the locus given by a circle.
00:00:38.560 --> 00:00:46.480
The equations, the modulus of π§ minus π§ one equals π, represent a circle with a radius of π and a centre at π§ one.
00:00:47.450 --> 00:00:57.720
The equations, the modulus of π§ minus π§ one equals π times the modulus of π§ minus π§ two, also represent a circle when π is greater than zero but not equal to one.
00:00:58.370 --> 00:01:02.610
In these situations, the radius and centre will need to be found for each case.
00:01:03.210 --> 00:01:16.570
We also need to know the equations of the form the modulus of π§ minus π§ one equals the modulus of π§ minus π§ two give a perpendicular bisector of the line segment that joins π§ one to π§ two, as shown in this diagram.
00:01:17.280 --> 00:01:21.260
And finally, a half line is described using the argument.
00:01:21.570 --> 00:01:28.340
The equations, the argument of π§ minus π§ one equals π, represent a half line from, but not including, π§ one.
00:01:28.890 --> 00:01:35.890
This half line makes an angle of π with the horizontal half line that also extends from π§ one in the positive π₯-direction.
00:01:36.650 --> 00:01:52.060
Just as with regions in the Cartesian plane, weβll also need to use the fact that a strict inequality, thatβs greater than or less than, will be represented by a dashed line whereas a weak inequality, greater than or equal to or less than or equal to, will be represented by a solid line.
00:01:53.030 --> 00:01:59.720
Letβs begin by considering a simple example that uses some of these definitions and look at how to decide which regions are the ones required.
00:02:00.890 --> 00:02:11.200
Sketch on an Argand diagram the region represented by the argument of π§ plus three minus two π is greater than or equal to negative π by two and less than π by four.
00:02:12.130 --> 00:02:15.500
To sketch this region, weβll begin by considering the boundaries.
00:02:15.830 --> 00:02:25.610
They are given by the argument of π§ plus three minus two π is equal to negative π by two and the argument of π§ plus three minus two π is equal to π by four.
00:02:26.060 --> 00:02:28.300
Each of these represents a half line.
00:02:28.970 --> 00:02:33.760
We can rewrite π§ plus three minus two π by factoring negative one.
00:02:33.760 --> 00:02:36.860
And we get π§ minus negative three plus two π.
00:02:37.500 --> 00:02:42.790
The point that represents this complex number will have Cartesian coordinates negative three, two.
00:02:43.420 --> 00:02:50.040
And of course, we represent this with an open circle since we know that the locus of points doesnβt actually include this point.
00:02:50.670 --> 00:02:57.880
The first boundary is going to make an angle of negative π by two with the positive horizontal, measured in a counterclockwise direction.
00:02:58.200 --> 00:03:03.420
This is the same as measuring an angle of positive π by two in the clockwise direction.
00:03:03.890 --> 00:03:05.750
And this is a weak inequality.
00:03:05.750 --> 00:03:08.280
So we draw a solid line for this one as shown.
00:03:08.860 --> 00:03:16.440
The half line for our next boundary will make an angle of π by four radians with the positive horizontal, measured in a counterclockwise direction.
00:03:17.290 --> 00:03:19.360
This time, itβs a weak inequality.
00:03:19.520 --> 00:03:21.960
So we need to draw a dashed line as shown.
00:03:23.140 --> 00:03:28.210
Now that we have the boundaries for our region, we need to decide which side of the region weβre going to shade.
00:03:28.660 --> 00:03:39.220
Weβre interested in all the complex numbers such that the argument of π§ plus three minus two π is greater than or equal to negative π by two and less than π by four.
00:03:39.600 --> 00:03:42.920
Thatβs going to be the region that lies between these two half lines.
00:03:43.130 --> 00:03:44.640
So we shade this region.
00:03:45.240 --> 00:03:45.910
And weβre done.
00:03:45.910 --> 00:03:48.610
Weβve sketched the required region on an Argand diagram.
00:03:49.720 --> 00:03:52.360
In our next example, weβll consider a circular region.
00:03:53.290 --> 00:03:55.470
The figure shows a region in the complex plane.
00:03:55.760 --> 00:03:58.660
Write an algebraic description of the shaded region.
00:03:59.230 --> 00:04:01.410
We can clearly see that this is a circle.
00:04:01.700 --> 00:04:05.980
But there are two ways to describe the locus that forms a circle.
00:04:06.370 --> 00:04:14.070
They are the modulus of π§ minus π§ one equals π and the modulus of π§ minus π§ one equals π times the modulus of π§ minus π§ two.
00:04:14.680 --> 00:04:18.210
In this example, it makes much more sense to use the first form.
00:04:18.650 --> 00:04:28.260
In fact, we try to use this form when describing regions as itβs much more simple to find the centre and the radius then find two points whose distance to the circle are in constant ratio.
00:04:28.890 --> 00:04:33.500
We can see that the centre of our circle is represented by the complex number four plus π.
00:04:33.940 --> 00:04:36.880
The Cartesian coordinates of this point are four, one.
00:04:37.710 --> 00:04:45.240
And we could use the distance formula to calculate the radius with either zero, seven or zero, negative five as one of the other points.
00:04:45.840 --> 00:04:52.870
Alternatively, we can find the modulus of the difference between the complex number four plus π and either seven π or negative five π.
00:04:53.350 --> 00:04:54.490
Letβs use seven π.
00:04:54.970 --> 00:05:01.060
Seven π minus four plus π is the same as six π minus four or negative four plus six π.
00:05:01.230 --> 00:05:03.990
So we need to find the modulus of negative four plus six π.
00:05:04.450 --> 00:05:11.620
To find the modulus, we square the real and imaginary parts, find their sum, and then find the square root of this number.
00:05:11.890 --> 00:05:16.810
So thatβs the modulus of negative four squared plus six squared, which is two root 13.
00:05:17.320 --> 00:05:28.690
So we know that the boundary for our region, the circle, is described by the equation, the modulus of π§ minus four plus π, cause thatβs the centre, is equal to two root 13 since thatβs the radius.
00:05:29.020 --> 00:05:31.710
And we can distribute these parentheses and write it as shown.
00:05:32.210 --> 00:05:34.470
We do however need to consider the region.
00:05:34.770 --> 00:05:37.360
Itβs the region outside of the circle.
00:05:37.730 --> 00:05:43.710
Each point in the region is further away from the centre of the circle than the distance of the radius.
00:05:44.080 --> 00:05:47.750
Itβs also a solid line which means it represents a weak inequality.
00:05:48.110 --> 00:05:56.640
And we can therefore say that the region is represented by the modulus of π§ minus four minus π is greater than or equal to two root 13.
00:05:57.300 --> 00:06:02.060
Weβve looked so far at simple regions defined by half lines and circles.
00:06:02.810 --> 00:06:06.040
To define composite regions, however, we need to use set notation.
00:06:06.260 --> 00:06:08.320
Weβll briefly recall the ones weβre interested in.
00:06:09.210 --> 00:06:14.770
π΄ union π΅ is the set of all elements in π΄ or π΅ or both.
00:06:15.480 --> 00:06:19.430
π΄ intersection π΅ is the set of all elements that are in both π΄ and π΅.
00:06:19.460 --> 00:06:20.320
Itβs the overlap.
00:06:20.840 --> 00:06:22.850
π΄ dash is the complement of π΄.
00:06:23.110 --> 00:06:27.960
And thatβs the set of all elements that are not in π΄, as shown by this third Venn diagram.
00:06:28.590 --> 00:06:32.650
Weβre going to now have a look at how we can define composite regions using this notation.
00:06:33.750 --> 00:06:45.460
Describe the shaded region in the following figure algebraically in the form π΄ intersection π΅ intersection πΆ, where π΄ is the set of complex numbers where the imaginary part of π§ is less than π΄.
00:06:45.740 --> 00:06:52.120
π΅ is the set of complex numbers such that the modulus of π§ is less than or equal to the modulus of π§ minus π§ one.
00:06:52.660 --> 00:06:58.750
And πΆ is the set of complex numbers such that the modulus of π§ is less than or equal to the modulus of π§ minus π§ two.
00:06:59.280 --> 00:07:04.540
And π, which is a real number, and π§ one and π§ two, which are complex numbers, are constants to be found.
00:07:05.190 --> 00:07:08.330
We can see that the region is bounded by three lines.
00:07:08.880 --> 00:07:14.620
We have one horizontal line, thatβs L one, and two diagonal lines that Iβve labelled L two and L three.
00:07:15.160 --> 00:07:20.510
We can see that our horizontal line has an equation of the imaginary part of π§ is equal to two.
00:07:20.880 --> 00:07:23.710
Now, weβre interested in the region directly below this.
00:07:23.840 --> 00:07:26.790
And we can see that itβs represented by a dashed line.
00:07:26.790 --> 00:07:28.660
So itβs going to be a strict inequality.
00:07:29.530 --> 00:07:36.330
We can say then that π΄ is equal to the set of complex numbers, where the imaginary part of this complex number must be less than two.
00:07:36.950 --> 00:07:41.620
Letβs now consider the diagonal line that passes through the points three, zero and zero, three.
00:07:42.290 --> 00:07:50.390
Remember, we described diagonal lines in the complex plane as perpendicular bisectors of a line segment joining a point to the origin.
00:07:50.870 --> 00:07:53.390
And itβs possible here to do this by inspection.
00:07:53.830 --> 00:07:59.950
Notice that the line itself passes through two vertices of a square at three, zero and zero, three.
00:08:00.590 --> 00:08:07.500
This means the line it bisects must pass through the other two vertices at zero, zero and three, negative three.
00:08:08.000 --> 00:08:15.410
So our line is the perpendicular bisector of the line segment joining the point three, negative three to the origin.
00:08:16.240 --> 00:08:19.140
That represents the complex number three minus three π.
00:08:19.530 --> 00:08:29.380
So we can say that π΅ is equal to the set of complex numbers such that the modulus of this complex number is less than or equal to the modulus of π§ minus three minus three π.
00:08:30.040 --> 00:08:31.670
Weβll now consider the third line.
00:08:32.000 --> 00:08:35.260
Itβs not quite so easy to find the line it bisects by inspection.
00:08:35.450 --> 00:08:37.990
So we can look at a method that generalises nicely.
00:08:38.370 --> 00:08:41.520
Weβll begin by finding the gradient of our line.
00:08:42.080 --> 00:08:47.940
The formula here is change in π¦ divided by change in π₯ or π¦ two minus π¦ one over π₯ two minus π₯ one.
00:08:48.490 --> 00:08:55.300
The gradient is therefore negative three minus zero over zero minus negative two, which is negative three over two.
00:08:55.850 --> 00:09:03.310
Since this line passes through the imaginary axis at negative three, we can say its equation is π¦ equals negative three over two π₯ minus three.
00:09:04.080 --> 00:09:07.680
And we can also find the gradient of the line it bisects.
00:09:08.120 --> 00:09:13.590
Since the line it bisects is perpendicular to this line, its gradient is two-thirds.
00:09:13.940 --> 00:09:15.760
And of course, it passes through the origin.
00:09:15.890 --> 00:09:18.330
So its equation is π¦ equals two-thirds π₯.
00:09:18.910 --> 00:09:21.370
Next, weβll find the point of intersection of these lines.
00:09:21.370 --> 00:09:22.790
And weβll do that by equating them.
00:09:23.480 --> 00:09:28.300
We add negative three over two π₯ to both sides and then divide through by three.
00:09:28.510 --> 00:09:33.560
And we see that the π₯-value of the point where these two lines intersect is negative 18 over 13.
00:09:33.940 --> 00:09:39.970
Substituting this into either of the equations we just used and we get a π¦-value of negative 12 over 13.
00:09:40.280 --> 00:09:43.200
So we now know the point at which these two lines intersect.
00:09:43.700 --> 00:09:50.770
Our line is the perpendicular line bisector of the line segment that joins a complex number with the origin.
00:09:51.150 --> 00:09:55.490
So this coordinate we just found must be half the value of the complex number we need.
00:09:55.880 --> 00:09:59.260
We can therefore double both the π₯- and π¦-values.
00:09:59.520 --> 00:10:09.910
And we can see that the complex number we need, the π§ two in part πΆ, is given by the point whose Cartesian coordinates are negative 36 over 13, negative 24 over 13.
00:10:10.520 --> 00:10:23.960
And that tells us that πΆ is equal to the set of complex numbers such that the modulus of π§ is less than or equal to the modulus of π§ minus negative 36 over 13 minus 24 over 13 π.
00:10:24.610 --> 00:10:28.960
And of course, the question did say that π, π§ one, and π§ two are constants to be found.
00:10:29.340 --> 00:10:31.270
So we add that π is equal to two.
00:10:31.550 --> 00:10:33.570
π§ is equal to three minus three π.
00:10:33.920 --> 00:10:39.000
And π§ two is equal to negative 36 over 13 minus 24 over 13.
00:10:39.750 --> 00:10:44.080
Weβll now consider one further example of representing composite regions in the complex plane.
00:10:44.870 --> 00:10:47.990
The complex number π§ satisfies the following conditions.
00:10:48.280 --> 00:10:53.500
The modulus of π§ is greater than or equal to two times the modulus of π§ plus 12 minus nine π.
00:10:54.440 --> 00:10:59.730
The modulus of π§ minus two π is greater than or equal to the modulus of π§ plus six plus four π.
00:11:00.300 --> 00:11:02.670
And the imaginary part of π§ is less than 12.
00:11:03.180 --> 00:11:05.220
Represent the region on an Argand diagram.
00:11:05.810 --> 00:11:07.770
Weβll begin by considering the first region.
00:11:08.550 --> 00:11:15.580
We can find the centre and radius of the circle by substituting π§ equals π₯ plus π¦π into our equation.
00:11:15.900 --> 00:11:19.070
Remember at the moment, weβre just finding the boundary for the region.
00:11:19.440 --> 00:11:22.180
We then square both sides of this equation.
00:11:22.530 --> 00:11:25.160
We can instantly replace two squared with four.
00:11:25.770 --> 00:11:28.700
But for the other bit, weβre going to need to use the definition of the modulus.
00:11:29.050 --> 00:11:34.240
We know that the modulus of π₯ plus π¦π is equal to the square root of π₯ squared plus π¦ squared.
00:11:34.550 --> 00:11:38.610
So the left-hand side of our equation becomes π₯ squared plus π¦ squared.
00:11:38.880 --> 00:11:41.630
We then gather the real and imaginary parts on the right-hand side.
00:11:41.740 --> 00:11:46.060
And we get four times π₯ plus 12 all squared plus π¦ minus nine all squared.
00:11:46.680 --> 00:11:56.450
Expanding the parentheses and then simplifying, on the right-hand side, we get four π₯ squared plus 96π₯ plus four π¦ squared minus 72π¦ plus 900.
00:11:56.810 --> 00:12:00.590
We subtract π₯ squared and π¦ squared from both sides of this equation.
00:12:01.050 --> 00:12:02.580
And then, we divide through by three.
00:12:03.110 --> 00:12:06.270
Now weβre looking to find the Cartesian equation of a circle.
00:12:06.450 --> 00:12:08.750
So weβre going to complete the square in π₯ and π¦.
00:12:09.160 --> 00:12:13.310
For π₯, we get π₯ plus 16 all squared minus 256.
00:12:13.720 --> 00:12:17.950
And for π¦, we get π¦ minus 12 all squared minus 144.
00:12:18.400 --> 00:12:20.240
And we add 300 of course.
00:12:20.550 --> 00:12:26.600
Negative 256 minus 144 plus 300 is negative 100.
00:12:26.860 --> 00:12:29.040
So we add 100 to both sides.
00:12:29.210 --> 00:12:32.120
And we have the Cartesian equation of a circle.
00:12:32.370 --> 00:12:37.130
It has a centre at negative 16, 12 and a radius of 10 units.
00:12:38.180 --> 00:12:43.030
The boundary for our first region is therefore this circle as shown.
00:12:43.620 --> 00:12:47.300
But how do we decide whether to shade inside or outside of this circle?
00:12:47.840 --> 00:12:51.080
Well, letβs choose a point that we know to be outside of the circle.
00:12:51.340 --> 00:12:54.230
Letβs choose the point whose Cartesian coordinates are one, zero.
00:12:54.810 --> 00:12:56.670
This is the complex number one.
00:12:57.130 --> 00:13:01.400
Weβre going to substitute this into the first inequality and see if the statement makes sense.
00:13:01.600 --> 00:13:08.220
This statement is the modulus of one is greater than or equal to two times the modulus of one plus 12 minus nine π.
00:13:08.640 --> 00:13:14.030
Or the modulus of one is greater than or equal to two times the modulus of 13 minus nine π.
00:13:14.310 --> 00:13:16.060
Well, the modulus of one is one.
00:13:16.550 --> 00:13:20.860
And the modulus of 13 minus nine π is root 250.
00:13:21.170 --> 00:13:25.070
Well, itβs not true that one is greater than two root 250.
00:13:25.220 --> 00:13:26.660
So this statement is false.
00:13:26.980 --> 00:13:30.790
And that tells us weβre going to be interested in the inside of the circle.
00:13:30.990 --> 00:13:33.730
Thatβs the region that satisfies that first condition.
00:13:34.200 --> 00:13:37.810
Weβll fully shade a region when weβve considered the other two situations.
00:13:37.990 --> 00:13:52.740
Now for two, we know that the equation, the modulus of π§ minus two π equals the modulus of π§ plus six plus four π, represents the perpendicular bisector of the line segment joining the point which represents two π and negative six minus four π.
00:13:53.180 --> 00:13:58.180
Thatβs the line segment between zero, two and negative six, negative four.
00:13:58.660 --> 00:14:07.740
We could find the exact equation of the perpendicular bisector of this line segment by considering the gradient and midpoint of the line segment it bisects.
00:14:08.110 --> 00:14:11.930
Alternatively, in this example, we can do this by inspection.
00:14:12.350 --> 00:14:18.470
And we can see that the line passes through the point zero, negative four and negative four, zero.
00:14:18.770 --> 00:14:21.690
And in fact, it also passes through the centre of our circle.
00:14:21.930 --> 00:14:26.530
Once again, weβll substitute π§ equals one into the inequality and see if the statement makes sense.
00:14:27.270 --> 00:14:30.560
The modulus of one minus two π is the square root of five.
00:14:30.790 --> 00:14:36.970
And the modulus of one plus six plus four π or the modulus of seven plus four π is the square root of 65.
00:14:37.210 --> 00:14:42.740
Once again, we can see that itβs actually not true that the square root of five is greater than or equal to the square root of 65.
00:14:43.660 --> 00:14:46.370
And we can see that weβre interested in the other side of the line.
00:14:46.940 --> 00:14:51.940
And of course, remember, we drew a solid line because our inequality is a weak inequality.
00:14:52.180 --> 00:14:53.840
Letβs now consider the third region.
00:14:54.180 --> 00:14:57.220
Weβre told that the imaginary part of π§ must be less than 12.
00:14:57.510 --> 00:15:03.090
The boundary of this region is the horizontal line passing through 12 on the imaginary axis.
00:15:03.360 --> 00:15:06.100
And because itβs a strict inequality, we draw a dash line.
00:15:06.640 --> 00:15:09.610
Weβre interested in the region below this dash line.
00:15:10.090 --> 00:15:15.610
To satisfy all three regions in this question, we need the intersection of the regions.
00:15:16.170 --> 00:15:19.600
So we shade the overlap between the three regions.
00:15:19.760 --> 00:15:21.710
Itβs the sector of the circle shown.
00:15:22.280 --> 00:15:23.150
And we are done.
00:15:23.150 --> 00:15:25.560
We have represented the region on an Argand diagram.
00:15:26.610 --> 00:15:31.530
In this video, weβve seen that we can use our understanding of loci to represent regions in the complex plane.
00:15:31.960 --> 00:15:39.460
We used dashed lines to represent boundary points that are not included and solid lines to represent regions that include their boundary points.
00:15:39.760 --> 00:15:46.950
And we can use set operations such as unions, intersections, and complements to define compound regions in the complex plane.