Video: Regions of the Complex Plane

In this video, we will learn how to express regions in the complex plane in terms of inequalities and how to interpret inequalities in terms of regions in the complex plane.

15:49

Video Transcript

In this video, we’ll learn how to express regions in a complex plane in terms of inequalities and how to interpret inequalities as regions in the complex plane. It’s likely you all have worked extensively with regions in the real Cartesian plane, considering them as inequalities. And this video looks to extend those concepts. We’ll 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.

Remember, there are two equations that are used to define the locus given by a circle. The equations, the modulus of 𝑧 minus 𝑧 one equals π‘Ÿ, represent a circle with a radius of π‘Ÿ and a centre at 𝑧 one. The equations, the modulus of 𝑧 minus 𝑧 one equals π‘˜ times the modulus of 𝑧 minus 𝑧 two, also represent a circle when π‘˜ is greater than zero but not equal to one. In these situations, the radius and centre will need to be found for each case.

We also need to know the equations of the form the modulus of 𝑧 minus 𝑧 one equals the modulus of 𝑧 minus 𝑧 two give a perpendicular bisector of the line segment that joins 𝑧 one to 𝑧 two, as shown in this diagram. And finally, a half line is described using the argument. The equations, the argument of 𝑧 minus 𝑧 one equals πœƒ, represent a half line from, but not including, 𝑧 one. This half line makes an angle of πœƒ with the horizontal half line that also extends from 𝑧 one in the positive π‘₯-direction.

Just as with regions in the Cartesian plane, we’ll also need to use the fact that a strict inequality, that’s 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. Let’s begin by considering a simple example that uses some of these definitions and look at how to decide which regions are the ones required.

Sketch on an Argand diagram the region represented by the argument of 𝑧 plus three minus two 𝑖 is greater than or equal to negative πœ‹ by two and less than πœ‹ by four.

To sketch this region, we’ll begin by considering the boundaries. They are given by the argument of 𝑧 plus three minus two 𝑖 is equal to negative πœ‹ by two and the argument of 𝑧 plus three minus two 𝑖 is equal to πœ‹ by four. Each of these represents a half line. We can rewrite 𝑧 plus three minus two 𝑖 by factoring negative one. And we get 𝑧 minus negative three plus two 𝑖. The point that represents this complex number will have Cartesian coordinates negative three, two. And of course, we represent this with an open circle since we know that the locus of points doesn’t actually include this point.

The first boundary is going to make an angle of negative πœ‹ by two with the positive horizontal, measured in a counterclockwise direction. This is the same as measuring an angle of positive πœ‹ by two in the clockwise direction. And this is a weak inequality. So we draw a solid line for this one as shown. The half line for our next boundary will make an angle of πœ‹ by four radians with the positive horizontal, measured in a counterclockwise direction. This time, it’s a weak inequality. So we need to draw a dashed line as shown.

Now that we have the boundaries for our region, we need to decide which side of the region we’re going to shade. We’re interested in all the complex numbers such that the argument of 𝑧 plus three minus two 𝑖 is greater than or equal to negative πœ‹ by two and less than πœ‹ by four. That’s going to be the region that lies between these two half lines. So we shade this region. And we’re done. We’ve sketched the required region on an Argand diagram.

In our next example, we’ll consider a circular region.

The figure shows a region in the complex plane. Write an algebraic description of the shaded region.

We can clearly see that this is a circle. But there are two ways to describe the locus that forms a circle. They are the modulus of 𝑧 minus 𝑧 one equals π‘Ÿ and the modulus of 𝑧 minus 𝑧 one equals π‘˜ times the modulus of 𝑧 minus 𝑧 two. In this example, it makes much more sense to use the first form. In fact, we try to use this form when describing regions as it’s much more simple to find the centre and the radius then find two points whose distance to the circle are in constant ratio.

We can see that the centre of our circle is represented by the complex number four plus 𝑖. The Cartesian coordinates of this point are four, one. 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. Alternatively, we can find the modulus of the difference between the complex number four plus 𝑖 and either seven 𝑖 or negative five 𝑖. Let’s use seven 𝑖.

Seven 𝑖 minus four plus 𝑖 is the same as six 𝑖 minus four or negative four plus six 𝑖. So we need to find the modulus of negative four plus six 𝑖. To find the modulus, we square the real and imaginary parts, find their sum, and then find the square root of this number. So that’s the modulus of negative four squared plus six squared, which is two root 13. So we know that the boundary for our region, the circle, is described by the equation, the modulus of 𝑧 minus four plus 𝑖, cause that’s the centre, is equal to two root 13 since that’s the radius. And we can distribute these parentheses and write it as shown.

We do however need to consider the region. It’s the region outside of the circle. Each point in the region is further away from the centre of the circle than the distance of the radius. It’s also a solid line which means it represents a weak inequality. And we can therefore say that the region is represented by the modulus of 𝑧 minus four minus 𝑖 is greater than or equal to two root 13.

We’ve looked so far at simple regions defined by half lines and circles. To define composite regions, however, we need to use set notation. We’ll briefly recall the ones we’re interested in.

𝐴 union 𝐡 is the set of all elements in 𝐴 or 𝐡 or both. 𝐴 intersection 𝐡 is the set of all elements that are in both 𝐴 and 𝐡. It’s the overlap. 𝐴 dash is the complement of 𝐴. And that’s the set of all elements that are not in 𝐴, as shown by this third Venn diagram. We’re going to now have a look at how we can define composite regions using this notation.

Describe the shaded region in the following figure algebraically in the form 𝐴 intersection 𝐡 intersection 𝐢, where 𝐴 is the set of complex numbers where the imaginary part of 𝑧 is less than 𝐴. 𝐡 is the set of complex numbers such that the modulus of 𝑧 is less than or equal to the modulus of 𝑧 minus 𝑧 one. And 𝐢 is the set of complex numbers such that the modulus of 𝑧 is less than or equal to the modulus of 𝑧 minus 𝑧 two. And π‘Ž, which is a real number, and 𝑧 one and 𝑧 two, which are complex numbers, are constants to be found.

We can see that the region is bounded by three lines. We have one horizontal line, that’s L one, and two diagonal lines that I’ve labelled L two and L three. We can see that our horizontal line has an equation of the imaginary part of 𝑧 is equal to two. Now, we’re interested in the region directly below this. And we can see that it’s represented by a dashed line. So it’s going to be a strict inequality. We can say then that 𝐴 is equal to the set of complex numbers, where the imaginary part of this complex number must be less than two.

Let’s now consider the diagonal line that passes through the points three, zero and zero, three. Remember, we described diagonal lines in the complex plane as perpendicular bisectors of a line segment joining a point to the origin. And it’s possible here to do this by inspection. Notice that the line itself passes through two vertices of a square at three, zero and zero, three. This means the line it bisects must pass through the other two vertices at zero, zero and three, negative three.

So our line is the perpendicular bisector of the line segment joining the point three, negative three to the origin. That represents the complex number three minus three 𝑖. So we can say that 𝐡 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 𝑧 minus three minus three 𝑖.

We’ll now consider the third line. It’s not quite so easy to find the line it bisects by inspection. So we can look at a method that generalises nicely. We’ll begin by finding the gradient of our line. The formula here is change in 𝑦 divided by change in π‘₯ or 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. The gradient is therefore negative three minus zero over zero minus negative two, which is negative three over two.

Since this line passes through the imaginary axis at negative three, we can say its equation is 𝑦 equals negative three over two π‘₯ minus three. And we can also find the gradient of the line it bisects. Since the line it bisects is perpendicular to this line, its gradient is two-thirds. And of course, it passes through the origin. So its equation is 𝑦 equals two-thirds π‘₯.

Next, we’ll find the point of intersection of these lines. And we’ll do that by equating them. We add negative three over two π‘₯ to both sides and then divide through by three. And we see that the π‘₯-value of the point where these two lines intersect is negative 18 over 13. Substituting this into either of the equations we just used and we get a 𝑦-value of negative 12 over 13. So we now know the point at which these two lines intersect. Our line is the perpendicular line bisector of the line segment that joins a complex number with the origin. So this coordinate we just found must be half the value of the complex number we need.

We can therefore double both the π‘₯- and 𝑦-values. And we can see that the complex number we need, the 𝑧 two in part 𝐢, is given by the point whose Cartesian coordinates are negative 36 over 13, negative 24 over 13. And that tells us that 𝐢 is equal to the set of complex numbers such that the modulus of 𝑧 is less than or equal to the modulus of 𝑧 minus negative 36 over 13 minus 24 over 13 𝑖. And of course, the question did say that π‘Ž, 𝑧 one, and 𝑧 two are constants to be found. So we add that π‘Ž is equal to two. 𝑧 is equal to three minus three 𝑖. And 𝑧 two is equal to negative 36 over 13 minus 24 over 13.

We’ll now consider one further example of representing composite regions in the complex plane.

The complex number 𝑧 satisfies the following conditions. The modulus of 𝑧 is greater than or equal to two times the modulus of 𝑧 plus 12 minus nine 𝑖. The modulus of 𝑧 minus two 𝑖 is greater than or equal to the modulus of 𝑧 plus six plus four 𝑖. And the imaginary part of 𝑧 is less than 12. Represent the region on an Argand diagram.

We’ll begin by considering the first region. We can find the centre and radius of the circle by substituting 𝑧 equals π‘₯ plus 𝑦𝑖 into our equation. Remember at the moment, we’re just finding the boundary for the region. We then square both sides of this equation. We can instantly replace two squared with four.

But for the other bit, we’re going to need to use the definition of the modulus. We know that the modulus of π‘₯ plus 𝑦𝑖 is equal to the square root of π‘₯ squared plus 𝑦 squared. So the left-hand side of our equation becomes π‘₯ squared plus 𝑦 squared. We then gather the real and imaginary parts on the right-hand side. And we get four times π‘₯ plus 12 all squared plus 𝑦 minus nine all squared. Expanding the parentheses and then simplifying, on the right-hand side, we get four π‘₯ squared plus 96π‘₯ plus four 𝑦 squared minus 72𝑦 plus 900. We subtract π‘₯ squared and 𝑦 squared from both sides of this equation. And then, we divide through by three.

Now we’re looking to find the Cartesian equation of a circle. So we’re going to complete the square in π‘₯ and 𝑦. For π‘₯, we get π‘₯ plus 16 all squared minus 256. And for 𝑦, we get 𝑦 minus 12 all squared minus 144. And we add 300 of course. Negative 256 minus 144 plus 300 is negative 100. So we add 100 to both sides. And we have the Cartesian equation of a circle. It has a centre at negative 16, 12 and a radius of 10 units.

The boundary for our first region is therefore this circle as shown. But how do we decide whether to shade inside or outside of this circle? Well, let’s choose a point that we know to be outside of the circle. Let’s choose the point whose Cartesian coordinates are one, zero. This is the complex number one. We’re going to substitute this into the first inequality and see if the statement makes sense.

This statement is the modulus of one is greater than or equal to two times the modulus of one plus 12 minus nine 𝑖. Or the modulus of one is greater than or equal to two times the modulus of 13 minus nine 𝑖. Well, the modulus of one is one. And the modulus of 13 minus nine 𝑖 is root 250. Well, it’s not true that one is greater than two root 250. So this statement is false. And that tells us we’re going to be interested in the inside of the circle. That’s the region that satisfies that first condition. We’ll fully shade a region when we’ve considered the other two situations.

Now for two, we know that the equation, the modulus of 𝑧 minus two 𝑖 equals the modulus of 𝑧 plus six plus four 𝑖, represents the perpendicular bisector of the line segment joining the point which represents two 𝑖 and negative six minus four 𝑖. That’s the line segment between zero, two and negative six, negative four. 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. Alternatively, in this example, we can do this by inspection. And we can see that the line passes through the point zero, negative four and negative four, zero. And in fact, it also passes through the centre of our circle.

Once again, we’ll substitute 𝑧 equals one into the inequality and see if the statement makes sense. The modulus of one minus two 𝑖 is the square root of five. And the modulus of one plus six plus four 𝑖 or the modulus of seven plus four 𝑖 is the square root of 65. Once again, we can see that it’s actually not true that the square root of five is greater than or equal to the square root of 65. And we can see that we’re interested in the other side of the line. And of course, remember, we drew a solid line because our inequality is a weak inequality.

Let’s now consider the third region. We’re told that the imaginary part of 𝑧 must be less than 12. The boundary of this region is the horizontal line passing through 12 on the imaginary axis. And because it’s a strict inequality, we draw a dash line. We’re interested in the region below this dash line. To satisfy all three regions in this question, we need the intersection of the regions. So we shade the overlap between the three regions. It’s the sector of the circle shown. And we are done. We have represented the region on an Argand diagram.

In this video, we’ve seen that we can use our understanding of loci to represent regions in the complex plane. We used dashed lines to represent boundary points that are not included and solid lines to represent regions that include their boundary points. And we can use set operations such as unions, intersections, and complements to define compound regions in the complex plane.

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