Lesson Explainer: Regions in the Complex Plane Mathematics

In this explainer, we will learn how to use loci to identify regions in the complex plane.

Before we work with regions in the complex plane, we will briefly recap some of the equations we use to define circles, lines, and half lines in the complex plane.

The types of regions we will consider in this explainer are the ones defined in terms of inequalities such as |π‘§βˆ’π‘§|<π‘ŸοŠ§ and composite regions defined in terms of multiple inequalities. For example, let us consider the region defined by |𝑧+1+𝑖|<2. To represent this on an Argand diagram, we first consider the circle defined by |𝑧+1+𝑖|=2. We notice that this is a circle of radius 2 centered at (βˆ’1,βˆ’1). Since the inequality is a strict inequality, we should draw this circle using a dashed line to represent the fact that the shaded region does not include its boundary.

Once we have drawn the boundary lines, we need to shade the region. To do this correctly, we need to consider whether we are to shade the interior or the exterior of the circle. This will be dictated by the direction of the inequality. If we are considering an inequality in the form of |π‘§βˆ’π‘§|<π‘ŸοŠ§, we will be concerned with the interior of the circle, whereas if we are considering |π‘§βˆ’π‘§|>π‘ŸοŠ§, we will be concerned with the exterior of the circle. The figure shows the regions in the complex plane represented by |𝑧+1+𝑖|<2 and |𝑧+1+𝑖|>2.

We will now demonstrate how we can recognize a region of the complex plane given to us in terms of an inequality.

Example 1: Representing Regions in the Complex Plane

Which of the following represents the region of the complex plane defined by βˆ’πœ‹2≀(𝑧+3βˆ’2𝑖)<πœ‹4?arg

Answer

We begin by considering the boundaries of the region. There are two we need to consider: arg(𝑧+3βˆ’2𝑖)=πœ‹4 and arg(𝑧+3βˆ’2𝑖)=βˆ’πœ‹2. Starting with the first of these, we notice that it is a half line emanating from βˆ’3+2𝑖 which makes a positive angle of πœ‹4 with the positive horizontal. Similarly, the second boundary of the region is defined by arg(𝑧+3βˆ’2𝑖)=βˆ’πœ‹2, which is a half line radiating from βˆ’3+2𝑖, which makes a negative angle of πœ‹2 with the positive horizontal. All of the figures include both of these boundary lines.

The region is defined by βˆ’πœ‹2≀(𝑧+3βˆ’2𝑖)<πœ‹4arg; therefore, it will be the region between these two half lines. This rules out both (a) and (e) as possible answers.

We now consider whether the boundary points are included in the region or not. Since arg(𝑧+3βˆ’2𝑖)<πœ‹4 is a strict inequality, this boundary of the region should be represented with a dashed line. This rules out (c) as a possible answer.

We are left with either (b) or (d) as possible answers. The difference between these two figures is whether we have used a solid circle β€’ or a hollow circle ∘ to represent the end point βˆ’3+2𝑖. Recall that we use the solid circle to indicate that the line includes the end point, whereas the hollow circle represents the fact that the end point is not included. Furthermore, recall that the argument of a complex number is not defined for zero.

Therefore, the region defined by βˆ’πœ‹2≀(𝑧+3βˆ’2𝑖)<πœ‹4arg will not include the end point and hence the correct representation of the region is (b).

Example 2: Describing Regions in the Complex Plane

The figure shows a region in the complex plane.

Write an algebraic description of the shaded region.

Answer

There is more than one way to describe a circle in the complex plane. For example, we could try to use the form |π‘§βˆ’π‘§|=π‘˜|π‘§βˆ’π‘§|. However, in most cases, it is a significantly more complicated process to try to find two points whose distance to the given circle are in constant ratio. Therefore, we tend to express regions involving circles using the form |π‘§βˆ’π‘§|=π‘ŸοŠ§. This is generally much easier to do since all we need to do is identify the center and radius.

In the figure, the center has been given to us. Therefore, we can simply state that 𝑧=4+π‘–οŠ§. However, we need to calculate the radius. Firstly, we notice that the circle intersects the imaginary axis at 7𝑖. Therefore, to find the radius, we can either use the Pythagorean theorem or simply evaluate the modulus of the difference of this complex number with π‘§οŠ§.

Hence, π‘Ÿ=|7π‘–βˆ’(4+𝑖)|=|βˆ’4+6𝑖|.

Using the definition of the modulus, we have π‘Ÿ=√(βˆ’4)+6=√52=2√13.

Notice that we are considering the region exterior to this circle. Therefore, we are considering the points which are further from the center than the radius. Furthermore, a solid line has been used in the figure which represents that the boundary points are included. Hence, the region is described by the following inequality: |π‘§βˆ’4βˆ’π‘–|β‰₯2√13.

To describe more interesting regions in the complex plane, we often want to talk about the regions that are subject to multiple constraints on the value of 𝑧. One of the common ways to do this is by using set notation and set operations.

Recall that the notation {π‘§βˆˆβ„‚βˆΆ(𝑧)<0}Re means that we are talking about the set of all complex numbers which have a positive real part. This notation is referred to as set builder notation. Furthermore, when talking about sets, there are a number of common operations we perform.

Unions, Intersections, and Complements

For two sets 𝐴,π΅βŠ‚π‘ˆ, where π‘ˆ is the universal set which contains both 𝐴 and 𝐡, we define the union of 𝐴 and 𝐡 to be the set of all elements that are in 𝐴 or in 𝐡. Notice that when we say β€œor” we are using the inclusive or; that is, we mean that an element is in 𝐴 or 𝐡 or in both 𝐴 and 𝐡. We use the notation 𝐴βˆͺ𝐡 to represent the union. Using set builder notation, we can write 𝐴βˆͺ𝐡={π‘₯βˆˆπ‘ˆβˆΆπ‘₯∈𝐴π‘₯∈𝐡}.or

The intersection of 𝐴 and 𝐡 is defined as the set of all elements that are in both 𝐴 and 𝐡. We write this as 𝐴∩𝐡. Using set builder notation, we can write 𝐴∩𝐡={π‘₯βˆˆπ‘ˆβˆΆπ‘₯∈𝐴π‘₯∈𝐡}.and

Finally, we define the complement of a set 𝐴 to be all of the elements that are in the universal set but not in 𝐴. We use the notation 𝐴 to represent this. Using set builder notation, we can write this as 𝐴={π‘₯βˆˆπ‘ˆβˆΆπ‘₯βˆ‰π΄}.

We can represent each one of these using a Venn diagram as follows.

Example 3: Representing Composite Regions in the Complex Plane

We define the regions 𝐴, 𝐡, and 𝐢 in the complex plane as 𝐴={π‘§βˆˆβ„‚βˆΆ(𝑧)<4},𝐡={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’8βˆ’12𝑖|},𝐢={π‘§βˆˆβ„‚βˆΆ|π‘§βˆ’6βˆ’5𝑖|<5}.Re

Which of the following figures could represent the region of the complex plane defined by (𝐴∩𝐡)βˆͺ𝐢?

Answer

The best approach to solving this problem is first to consider the regions represented by 𝐴, 𝐡, and 𝐢 separately. Beginning with 𝐴, we will first consider the region represented by 𝐴={π‘§βˆˆβ„‚βˆΆ(𝑧)<4}Re and then its complement. Notice that 𝐴 represents all the complex numbers with real parts less than 4. Hence, this will be the region to the left of the line π‘₯=4. Notice that since the inequality is strict, we will represent the boundary with a dashed line. The figure below shows the region 𝐴.

We, however, do not need to consider the region 𝐴; we need to consider 𝐴. This will be the region of all the points not in 𝐴. We can think about this visually, by inverting the regions we have shaded in the plane. Notice, when doing this, that we need to change the boundary from a dashed line to a solid line. Alternatively, we can think about this more abstractly as follows. Since 𝐴 represents all complex numbers with a real part less than 4, 𝐴 will represent all complex numbers with a real part greater than or equal to 4. Whichever approach we use, we can represent 𝐴 as the following region.

We now consider 𝐡={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’8βˆ’12𝑖|}. We start by considering the boundary of the region where |𝑧|≀|π‘§βˆ’8βˆ’12𝑖|. Notice that this represents the perpendicular bisector of the line segment between the origin and the point represented by 𝑧=8+12π‘–οŠ§. We can draw this on an Argand diagram with a solid line since the region contains the boundary points. We then notice that the inequality represents all points that are closer to the origin than 8+12𝑖. Hence, we can represent the region 𝐡 as follows.

We can now consider the intersection 𝐴∩𝐡 of these two regions. This will be all of the points that are in both of the regions. We can represent this as follows.

We can now consider 𝐢={π‘§βˆˆβ„‚βˆΆ|π‘§βˆ’6βˆ’5𝑖|<5}. This represents a circle of radius 5 centered at 6+5𝑖. Notice that since the inequality is strict, the region does not contain the boundary. Hence, we can represent this region as follows.

Finally, we consider the union of 𝐴∩𝐡 and 𝐢. Recall that this is the region we get by adding all the points in both regions together. We can represent this as follows.

Hence, the correct answer is (d).

Example 4: Describing Composite Regions in the Complex Plane

The shaded region in the following figure can be algebraically described by 𝐴∩𝐡∩𝐢, where 𝐴={π‘§βˆˆβ„‚βˆΆ(𝑧)<π‘Ž},𝐡={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’π‘§|},𝐢={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’π‘§|}.Im

Find the values of π‘Ž, π‘§οŠ§, and π‘§οŠ¨, where π‘Žβˆˆβ„ and 𝑧,π‘§βˆˆβ„‚οŠ§οŠ¨.

Answer

The region is bounded by three lines: 𝐿, 𝐿, and 𝐿 as shown below.

The first of these, 𝐿, is represented by Im(𝑧)=2. The region does not contain the boundary points along this line. Hence, 𝐴={π‘§βˆˆβ„‚βˆΆ(𝑧)<2}Im. We now turn our attention to 𝐿. We need to describe this line as a perpendicular bisector of a line segment between the origin and a another point which we have labeled π‘ƒοŠ§. Sometimes, as in this example, it is possible to do this by inspection. We can see that the line passes through two of the diagonally opposite vertices of the square defined by the points (0,0), (3,0), (3,βˆ’3), and (0,βˆ’3). Therefore, it is the perpendicular bisector of the line segment between the other two diagonally opposite vertices. Hence, 𝐿 is the perpendicular bisector of the line segment between (0,0) and 𝑃(3,βˆ’3).

Therefore, 𝐡={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’(3βˆ’3𝑖)|}. Finally, we turn our attention to the third line, 𝐿. Once again, we need to describe this line as a perpendicular bisector of a line segment between the origin and a another point which we have labeled π‘ƒοŠ¨. In this case, it is more difficult to do this by inspection. Therefore, we will demonstrate a technique that generalizes more easily.

We begin by finding the slope π‘š of the line 𝐿. We can calculate this using the formula π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

Since the line passes through the points (0,βˆ’3) and (βˆ’2,0), we have π‘š=βˆ’3βˆ’00βˆ’(βˆ’2)=βˆ’32.

We can therefore write the equation of the line as 𝑦=βˆ’32π‘₯βˆ’3.

We now find the equation of a line that is perpendicular to this which passes through the origin. Recall that the product of the slopes of two perpendicular lines is βˆ’1. Hence, the slope of a line which is perpendicular to 𝐿 is 23. Therefore, the equation of a perpendicular line that passes through the origin is given by 𝑦=23π‘₯.

We now need to find a point on this line whose distance to 𝐿 is equal to the distance between 𝐿 and the origin. To do this, we first find the point of intersection of these two lines. By equating the two equations of the lines, we can solve for π‘₯ as follows: 23π‘₯=βˆ’32π‘₯βˆ’3.

Gathering our π‘₯ terms on the left-hand side, we have 136π‘₯=βˆ’3.

Hence, π‘₯=βˆ’1813.

By substituting this back into one of the two equations, we find the corresponding value of 𝑦 to be 𝑦=βˆ’1213. We can represent this as a complex number 𝑀=βˆ’1813βˆ’1213𝑖. We would like this point to lie at the midpoint of π‘ƒοŠ¨ and the origin. Recall that we can find the midpoint of two complex numbers by simply taking their average. Hence, since 𝑀 is the midpoint of the line segment between the origin and the point π‘ƒοŠ¨ which we will represent with the complex number π‘§οŠ¨, we can write 𝑀=12(𝑧+0).

Hence, 𝑧=2𝑀=2ο€Όβˆ’1813βˆ’1213π‘–οˆ=βˆ’3613βˆ’2413𝑖.

Hence, 𝐢=ο¬π‘§βˆˆβ„‚βˆΆ|𝑧|≀|||π‘§βˆ’ο€Όβˆ’3613βˆ’2413π‘–οˆ|||. Therefore, we can represent the region as 𝐴∩𝐡∩𝐢, where 𝐴={π‘§βˆˆβ„‚βˆΆ(𝑧)<2},𝐡={π‘§βˆˆβ„‚βˆΆ|𝑧|≀|π‘§βˆ’(3βˆ’3𝑖)|},𝐢=ο¬π‘§βˆˆβ„‚βˆΆ|𝑧|≀|||π‘§βˆ’ο€Όβˆ’3613βˆ’2413π‘–οˆ|||.Im

Example 5: Drawing Composite Regions in the Complex Plane

The complex number 𝑧 satisfies the following conditions: |𝑧|β‰₯2|𝑧+12βˆ’9𝑖|,|π‘§βˆ’2𝑖|β‰₯|𝑧+6+4𝑖|,(𝑧)<12.Im

  1. Which of the following shaded regions represents the locus of 𝑧?
  2. Find the area of the locus of 𝑧.

Answer

Part 1

We consider the regions represented by each condition in order. Starting with the first condition, |𝑧|β‰₯2|𝑧+12βˆ’9𝑖|, its boundary is a circle in the complex plane. To find its center and origin, we can find its Cartesian equation by substituting 𝑧=π‘₯+𝑦𝑖 into the equation |𝑧|=2|𝑧+12βˆ’9𝑖| and rearranging it into the standard form of a circle. Hence, |π‘₯+𝑦𝑖|=2|π‘₯+𝑦𝑖+12βˆ’9𝑖|=2|(π‘₯+12)+(π‘¦βˆ’9)𝑖|.

Squaring both sides of the equation, we have |π‘₯+𝑦𝑖|=4|(π‘₯+12)+(π‘¦βˆ’9)𝑖|.

Using the definition of the modulus, we can rewrite this as π‘₯+𝑦=4ο€Ή(π‘₯+12)+(π‘¦βˆ’9).

Expanding the parentheses yields π‘₯+𝑦=4ο€Ήπ‘₯+24π‘₯+144+π‘¦βˆ’18𝑦+81=4π‘₯+96π‘₯+4π‘¦βˆ’72𝑦+900.

Gathering all the terms on the right-hand side of the equation, we have 0=3π‘₯+96π‘₯+3π‘¦βˆ’72𝑦+900.

Notice that there is a common factor of 3 in all the terms. Hence, we can cancel this as follows: 0=π‘₯+32π‘₯+π‘¦βˆ’24+300.

We now complete the square in π‘₯ and 𝑦 as follows: 0=(π‘₯+16)βˆ’256+(π‘¦βˆ’12)βˆ’144+300=(π‘₯+16)+(π‘¦βˆ’12)βˆ’100.

We can rewrite this as (π‘₯+16)+(π‘¦βˆ’12)=10 which is the equation of a circle of radius 10 centered at (βˆ’16,12). Therefore, we can represent the region of the complex plane 𝐴 represented by the inequality |𝑧|β‰₯2|𝑧+12βˆ’9𝑖| as follows.

We now consider the second condition |π‘§βˆ’2𝑖|β‰₯|𝑧+6+4𝑖|. The boundary of this region is the perpendicular bisector of the line segment between (0,2) and (βˆ’6,βˆ’4). Therefore, the region 𝐡 represented by the inequality |π‘§βˆ’2𝑖|β‰₯|𝑧+6+4𝑖| will be the half plane above this line.

The section of the plane which satisfies both of these conditions, |𝑧|β‰₯2|𝑧+12βˆ’9𝑖| and |π‘§βˆ’2𝑖|β‰₯|𝑧+6+4𝑖|, will be given by the intersection of these two regions as shown below.

Finally, we consider the region represented by the points satisfying the third condition: Im(𝑧)<12. This represents all of the complex numbers with an imaginary part less than 12. Hence, the region 𝐢 represented by the inequality Im(𝑧)<12 is the half plane below the line 𝑦=12. Notice that the equality is strict so we represent the boundary of this region by a dotted line.

Finally, we consider the region which satisfies all the conditions, |𝑧|β‰₯2|𝑧+12βˆ’9𝑖|, |π‘§βˆ’2𝑖|β‰₯|𝑧+6+4𝑖|, and Im(𝑧)<12, given by the intersection of these regions as shown below.

Part 2

To find the area of this region, we notice that it is a sector of the circle. Hence, its area will be given by 12π‘ŸπœƒοŠ¨, where π‘Ÿ is the radius of the circle and πœƒ is the angle subtended by the arc at the center of the circle.

The angle πœƒ at 𝐢 and ∠𝐢𝐷𝐴 are alternative angles and, hence, equal. Furthermore, by considering the triangle 𝐴𝐸𝐷, we can see πœƒ=πœ‹2βˆ’πœ™.

We can find the measure of πœ™ using the coordinates of points 𝐴 and 𝐡 as follows: πœ™=ο€½62βˆ’(βˆ’4)=πœ‹4.arctan

Therefore, πœƒ=πœ‹4 and the area of the region is given by 12π‘Ÿπœƒ=12ο€Ή10ο…ο€»πœ‹4=25πœ‹2.

Example 6: Composite Regions in the Complex Plane

Consider the following regions in the complex plane: 𝐴={π‘§βˆˆβ„‚βˆΆ2≀|𝑧|<4},𝐡={π‘§βˆˆβ„‚βˆΆ(𝑧)(𝑧)<0}.ReIm

The region 𝑅 is defined as 𝑅=𝐴∩𝐡.

  1. Which of the following shaded regions represents 𝑅 on an Argand diagram?
  2. Find the area of the region 𝑅.

Answer

Part 1

We begin by considering region 𝐴. It is defined as the region between the two circles |𝑧|=2 and |𝑧|=4. Notice that the region contains the inner boundary, but not the outer. Hence, we can represent this region as follows.

We now consider the region 𝐡. This is the region on the complex plane where the product of the real and imaginary parts is negative. For this condition to be satisfied, we need either the real part or the imaginary part to be negative, but not both. Hence, this region is the second and fourth quadrants of the plane. Once again, notice that this does not contain the boundary; therefore, we can represent this in an Argand diagram as follows.

Before we consider the region 𝑅, we first consider the complement of 𝐡. This will be the first and third quadrants of the complex plane. However, since 𝐡 did not include its boundary points, 𝐡 will. Therefore, we can represent 𝐡 in an Argand diagram as follows.

Finally, we would like to take the intersection of 𝐴 with 𝐡. This is the region of points in both 𝐴 and 𝐡. Hence, we can represent this as follows.

Part 2

To find the area of this region, we notice that it will be equal to half the area of the larger circle minus half the area of the smaller circle: 12πœ‹π‘Ÿβˆ’12πœ‹π‘Ÿ=12πœ‹ο€Ήπ‘Ÿβˆ’π‘Ÿο….

Substituting in the values of π‘ŸοŠ§ and π‘ŸοŠ¨, we have that the area of section 𝑅 is given by 12πœ‹ο€Ήπ‘Ÿβˆ’π‘Ÿο…=12πœ‹ο€Ή4βˆ’2=12πœ‹(12)=6πœ‹.

Key Points

  • Using our knowledge of loci in the complex plane, we can represent regions in the complex plane on an Argand diagram.
  • We use dashed lines to represent the boundary points that are not included and solid lines to represent regions that include their boundary points. Whether a given region contains its boundary points or not will be determined by whether the inequalities are strict or not.
  • We use set operations such as unions, intersections, and complements to define compound regions in the complex plane.

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