Explainer: Loci in the Complex Plane Using the Modulus

In this explainer, we will learn how to find the loci of a complex equation in the complex plane from the modulus.

The fact that complex numbers can be represented on an Argand Diagram furnishes them with a lavish geometry. This geometry is further enriched by the fact that we can consider complex numbers either as points in the plane or as vectors. Using both of these interpretations, we can develop a deep understanding of the geometric properties of transformations that are represented by certain complex operations. To lay the foundations to gaining a deeper understanding of these things, we turn our attention to understanding loci in the complex plane. The locus of a point 𝑧 is the set of all the points 𝑧 which satisfy a particular condition.

Recall that the modulus represents the distance of a point from the origin. For example, given the point 𝑀=65+8𝑖5, we can calculate the modulus as follows: |𝑀|=ο„Ÿο€Ό65+ο€½8𝑖5=√4=2.

Therefore, we know that the point 𝑀 is at a distance of 2 from the origin. Rather than asking what the modulus of a particular complex number is, we could ask the question β€œwhat is the locus of a point 𝑧 of fixed modulus, say |𝑧|=2?” This is equivalent to the locus of a point at a fixed distance of 2 from the origin. From elementary geometry, we know that this represents a circle of radius 2 centered at the origin.

Example 1: Geometry in the Complex Plane

A complex number 𝑀 lies at a distance of 5√2 from 𝑧=92+72π‘–οŠ§ and a distance of 4√5 from 𝑧=βˆ’92βˆ’72π‘–οŠ¨. Does the point 𝑀 lie on the circle centered at the origin that passes through π‘§οŠ§ and π‘§οŠ¨?

Answer

We would like to know whether the point 𝑀 lies on the circle centered at the origin which passes through π‘§οŠ§ and π‘§οŠ¨. To do this, we could try to calculate the possible positions of 𝑀 and then find its modulus. However, an easier approach would be to try to use the properties of circles to assess whether this 𝑀 lies on the circle. Firstly, we notice that since 𝑧=βˆ’π‘§οŠ§οŠ¨, they are diametrically opposite. Hence, the line segment from π‘§οŠ§ to π‘§οŠ¨ forms a diameter of the circle. Therefore, 𝑀 will lie on the circle if the triangle formed by 𝑀, π‘§οŠ§, and π‘§οŠ¨ is a right triangle.

To check whether this triangle is indeed a right triangle, we can use the reverse Pythagorean theorem to check whether the lengths of the sides π‘Ž, 𝑏, and 𝑐 satisfy π‘Ž+𝑏=π‘οŠ¨οŠ¨οŠ¨, where 𝑐 is the length of the longest side. We have already been given the length of two of the sides, so we only need to calculate the distance between the points π‘§οŠ§ and π‘§οŠ¨. Recall that we can find the distance between two complex numbers by evaluating the modulus of their difference. Hence, the distance between π‘§οŠ§ and π‘§οŠ¨ is given by 𝑑=|π‘§βˆ’π‘§|.

Substituting in the values of π‘§οŠ§ and π‘§οŠ¨, we have 𝑑=|||92+72π‘–βˆ’ο€Όβˆ’92βˆ’72π‘–οˆ|||=|9+7𝑖|.

Using the definition of the modulus, we have 𝑑=√9+7=√130.

We now can check whether these three side lengths satisfy the Pythagorean theorem. To do this, we first need to identify the longest side, which is the side between π‘§οŠ§ and π‘§οŠ¨. Hence, we need to verify whether π‘‘οŠ¨ is equal to the sum of the squares of the other two sides: ο€»5√2+ο€»4√5=50+80=130=𝑑.

Since the side lengths satisfy the Pythagorean theorem, we can conclude that the triangle is indeed a right triangle and hence that 𝑀 does lie on the circle passing through π‘§οŠ§ and π‘§οŠ¨.

This example demonstrates how using our knowledge of geometry will aid us significantly when working to understand the properties of loci in the complex plane. Secondly, it reminds us that the distance between two points is given by the modulus of their difference. This property of the modulus helps us interpret the locus of the point 𝑧 such that |π‘§βˆ’π‘§|=π‘Ÿ, where π‘§βˆˆβ„‚οŠ§ is constant. Given that π‘§οŠ§ is constant, this is the locus of the point 𝑧 which is a fixed distance of π‘Ÿ from π‘§οŠ§. Hence, this represents a circle centered at π‘§οŠ§ of radius π‘Ÿ.

Another way to understand this equation is by recalling that the geometric interpretation of addition is a translation. Therefore, to find the equation representing a circle of radius π‘Ÿ centered at π‘§οŠ§, we can first transform the point π‘§οŠ§ to the origin. This translation can be represented by subtracting π‘§οŠ§. Once we have made this translation, a circle of radius π‘Ÿ is the locus of a point with constant modulus π‘Ÿ. Hence, the equation representing a circle of radius π‘Ÿ centered at π‘§οŠ§ is given by |π‘§βˆ’π‘§|=π‘Ÿ.

The Loci of Points Forming Circles Centered at 𝑧₁ of Radius π‘Ÿ

For a given constant π‘§βˆˆβ„‚οŠ§, the locus of a point 𝑧, which satisfies |π‘§βˆ’π‘§|=π‘ŸοŠ§, is a circle centered at π‘§οŠ§ of radius π‘Ÿ.

Example 2: Working with Loci

A complex number 𝑧 satisfies |π‘§βˆ’2+3𝑖|=2.

  1. Describe the locus of 𝑧 and give its Cartesian equation.
  2. Find the range of the argument of 𝑧 in the range [βˆ’πœ‹,πœ‹], correct to three significant figures.
  3. Find the range of the modulus of 𝑧.

Answer

Part 1

Since 𝑧 satisfies |π‘§βˆ’2+3𝑖|=2, its locus represents the points of a constant distance of 2 from the complex number 𝑧=2βˆ’3π‘–οŠ§. Hence, it is a circle of radius 2 centered at 2βˆ’3𝑖.

To find the Cartesian equation, we can substitute 𝑧=π‘₯+𝑦𝑖 into the equation and then manipulate it into a standard form as we will demonstrate. Setting 𝑧=π‘₯+𝑦𝑖, we have |π‘₯+π‘¦π‘–βˆ’2+3𝑖|=2.

Gathering the real and imaginary parts, we have |π‘₯βˆ’2+(𝑦+3)𝑖|=2.

Now we can use the definition of the modulus to rewrite this as (π‘₯βˆ’2)+(𝑦+3)=2.

Squaring both sides of the equation yields (π‘₯βˆ’2)+(𝑦+3)=4, which is a Cartesian equation for a circle of radius 2 centered at (2,βˆ’3).

Part 2

Drawing the locus on an Argand diagram will be extremely useful to help us find the range of the argument of 𝑧.

We can clearly see that the minimum value of the argument is βˆ’πœ‹2. We denote the maximum value by πœƒβˆ’πœ‹2. Using the properties of circles, we can see that the triangles 𝑂𝐡𝐢 and 𝑂𝐴𝐢 are similar and, hence, the angles at 𝑂 are equal. Therefore, we can find πœƒ2 using triangle 𝑂𝐴𝐢 as follows: tanπœƒ2=𝐴𝐢𝑂𝐴=23.

Hence, πœƒ=2ο€Ό23.arctan

Therefore, the range of the argument of 𝑧 is ο”βˆ’πœ‹2,2ο€Ό23οˆβˆ’πœ‹2arctan. Evaluating these values, we can write that the range of 𝑧, correct to three significant figures, is [βˆ’1.57,βˆ’0.395].

Part 3

Finding the range of the modulus is somewhat easier. The maximum and minimum values will be on the half-line extending from 𝑂 to 𝐢 and beyond. We already know the radius of the circle π‘Ÿ. Hence, minimum and maximum values of |𝑧| will be given by |𝑂𝐢|βˆ’π‘Ÿ and |𝑂𝐢|+π‘Ÿ respectively. Since 𝐢 is the center of the circle represented by 𝑧=2βˆ’3π‘–οŠ§, we have that |𝑂𝐢|=|𝑧|=2+(βˆ’3)=√13. Therefore, the range of |𝑧| is given by ο“βˆš13βˆ’2,√13+2.

Using our knowledge of geometry, we can also find the equation of a given locus as the next example will demonstrate.

Example 3: Finding the Equation of a Locus

The figure shows a circular locus of a point 𝑧 in the complex plane. Given that 𝐢 is the centre of the circle, write an equation for the locus in the form |π‘§βˆ’π‘Ž|=𝑏, where π‘Žβˆˆπ‘ and 𝑏>0 are constants to be found.

Answer

The locus of the point 𝑧 is a circle to write its equation in the form |π‘§βˆ’π‘Ž|=𝑏, we need to find its center, represented by point π‘Ž, and its radius, represented by the real number 𝑏. From the figure, we can see that the locus passes through three points: 𝑧=0, 𝑧=4π‘–οŠ¨, and 𝑧=βˆ’10 which form a right triangle. Using the properties of circles, we know that the hypotenuse of this triangle will be the diameter of the circle. Hence, the radius is half the length of the hypotenuse. Therefore, 𝑏=12√10+4=12√116.

Simplifying, we have 𝑏=√29.

To find the center of the circle, we can use the fact that the midpoint of two complex numbers 𝑧 and 𝑀 is given by 12(𝑧+𝑀). Therefore, π‘Ž=βˆ’10+4𝑖2=βˆ’5+2𝑖.

Hence, the given figure is the locus of the point 𝑧 satisfying |π‘§βˆ’(βˆ’5+2𝑖)|=√29.

The next type of loci we will look at is ones defined by equations of the form |π‘§βˆ’π‘§|=|π‘§βˆ’π‘§|, where 𝑧,π‘§βˆˆβ„‚οŠ§οŠ¨ are constants. The left-hand side represents the distance of 𝑧 from the fixed point π‘§οŠ§, and likewise the right-hand side is the distance of the point 𝑧 from π‘§οŠ¨. Therefore, the equation represents the locus of a point which is equidistant from the two points π‘§οŠ§ and π‘§οŠ¨. Therefore, the locus is the perpendicular bisector of the line segment joining π‘§οŠ§ to π‘§οŠ¨ as shown in the figure.

The Locus of a Point Equidistant from Two Points

For two given constants 𝑧,π‘§βˆˆβ„‚οŠ§οŠ¨, the locus of a point 𝑧 which satisfies |π‘§βˆ’π‘§|=|π‘§βˆ’π‘§| is a the perpendicular bisector of the line segment joining π‘§οŠ§ to π‘§οŠ¨.

Example 4: Equations of a Locus

A complex number 𝑧 satisfies |𝑧+1+𝑖|=|π‘§βˆ’2βˆ’6𝑖|.

  1. Describe the locus of 𝑧 and give its Cartesian equation.
  2. What is the minimum value of |𝑧|?

Answer

Part 1

It is often helpful to put the equation into the standard |π‘§βˆ’π‘§|=|π‘§βˆ’π‘§| form to ensure we correctly identify π‘§οŠ§ and π‘§οŠ¨. Doing this yields |π‘§βˆ’(βˆ’1βˆ’π‘–)|=|π‘§βˆ’(2+6𝑖)|.

Therefore, the locus of 𝑧 is the perpendicular bisector of the line segment between βˆ’1βˆ’π‘– and 2+6𝑖.

To find its Cartesian equation, we first calculate the slope π‘šοŠ§ of the line segment joining π‘§οŠ§ to π‘§οŠ¨. From this, we will be able to find the slope of the line representing the locus of 𝑧.

Using the formula for the slope, π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯, we find that π‘š=6βˆ’(βˆ’1)2βˆ’(βˆ’1)=73.

We now need to find the slope representing the locus of 𝑧 which we denote π‘šοŠ¨. Since this is perpendicular to this line segment joining π‘§οŠ§ to π‘§οŠ¨, we can use the property that the product of the slopes of perpendicular lines is βˆ’1. Hence, we have π‘šπ‘š=βˆ’1.

Therefore, the slope π‘šοŠ¨ is given by π‘š=βˆ’1π‘š=βˆ’37.

Now that we have the slope of the line, we need to find a point that it passes through. Since the line represents all the points which are equidistant from the two points π‘§οŠ§ and π‘§οŠ¨, the line will pass through the midpoint of the line segment joining π‘§οŠ§ to π‘§οŠ¨. Recall that we can find the midpoint of two complex numbers by simply taking their average. Hence, the midpoint 𝑀 of the line segment is given by 𝑀=12(𝑧+𝑧)=12(βˆ’1βˆ’π‘–+2+6𝑖).

Simplifying, we get 𝑀=12(1+5𝑖)=12+52𝑖.

We can now use the point–slope form of a line, π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), to write the Cartesian equation of the locus of 𝑧. Substituting in the coordinates of the points representing 𝑀 and the slope of the line, we have π‘¦βˆ’52=βˆ’37ο€Όπ‘₯βˆ’12.

Rearranging, we find that 𝑦=βˆ’37π‘₯+197.

Part 2

To find the minimum value of |𝑧|, we are looking for the point 𝐢 on the locus of 𝑧 which is closest to the origin. The rules of elementary geometry tell us that the line segment joining 𝐢 to the origin will be perpendicular to the line which represents the locus of 𝑧. Hence, its slope will be the same as the slope of the line segment joining π‘§οŠ§ to π‘§οŠ¨.

This means that we can write a Cartesian equation for the line through the origin and 𝐢 as 𝑦=73π‘₯.

Therefore, 𝐢 lies at the point of interaction of this line with the line representing the locus of 𝑧: 𝑦=βˆ’37π‘₯+197.

By equating the values of 𝑦, we can solve for the π‘₯-coordinate of 𝐢 as follows: 73π‘₯=βˆ’37π‘₯+197.

Adding βˆ’37π‘₯ to both sides of the equation yields 5821π‘₯=197.

Hence, π‘₯=5758.

Substituting this into one of the equations of the lines gives us the value of 𝑦 as follows: 𝑦=73Γ—5758=13358.

Therefore, the minimum value of |𝑧| is given by |𝑧|=ο„Ÿο€Ό5758+ο€Ό13358=√3,249+17,68958.

Simplifying gives |𝑧|=√20,93858=1958√58.

Example 5: Finding the Equations of Loci

The figure shows a locus of a point 𝑧 in the complex plane. Write an equation for the locus in the form |𝑧+2+7𝑖|=|π‘§βˆ’π‘Ž|, where π‘Žβˆˆβ„‚ is a constant to be found.

Answer

Firstly, we consider the geometric interpretation of the equation |𝑧+2+7𝑖|=|π‘§βˆ’π‘Ž|. This equation represents the perpendicular bisector of the line segment from the points represented by π‘Ž and βˆ’2βˆ’7𝑖. We can represent this in a diagram as follows.

We will denote the points represented by the complex numbers π‘Ž, 𝑏, and 𝑐 as 𝐴, 𝐡, and 𝐢. The point 𝐢 represents the intersection of the line segment joining 𝐴 and 𝐡 which we will denote 𝐿 and the line representing the locus of 𝑧 which we will denote 𝐿. We can therefore find the point 𝐢 by finding the equations of these two lines. Starting with 𝐿, since it passes through the points (0,4) and (βˆ’3,0), we can find its slope as follows: π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯=4βˆ’00βˆ’(βˆ’3)=43.

Using the slope–intercept form, we can write the equation of the line 𝐿 as 𝑦=43π‘₯+4.

We now consider the equation of 𝐿. Since it is perpendicular to 𝐿, the product of their slopes will be negative one. Hence, π‘š=βˆ’1π‘š=βˆ’34.

Since this line passes through the point 𝐡(βˆ’2,βˆ’7), we can now use the slope–point form, π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), to write its equation as 𝑦+7=βˆ’34(π‘₯+2).

Subtracting 7 from both sides gives 𝑦=βˆ’34π‘₯βˆ’172.

By equating the equations for 𝐿 and 𝐿, we can now find the π‘₯-coordinate of 𝐢 as follows: 43π‘₯+4=βˆ’34π‘₯βˆ’172.

Adding βˆ’34π‘₯ and subtracting 4 from both sides of the equation yields 2512π‘₯=βˆ’252.

Hence, π‘₯=βˆ’6. Substituting this back into the equation for 𝐿 gives 𝑦=43(βˆ’6)+4, which simplifies to 𝑦=βˆ’4. Therefore, expressed as a complex number the point 𝐢 is 𝑐=βˆ’6βˆ’4𝑖.

Finally, to find the value of π‘Ž, we can use the interpretation of complex numbers as vectors. The complex number that represents the vector from 𝐡 to 𝐢, can be written as βˆ’π‘+𝑐. Hence, the complex number which represents the vector from 𝐡 to 𝐴 is given by 2(βˆ’π‘+𝑐). Adding this vector to 𝑏 will give us the complex number π‘Ž. Hence, π‘Ž=𝑏+2(βˆ’π‘+𝑐)=2π‘βˆ’π‘.

Substituting in the complex numbers 𝑏 and 𝑐 gives π‘Ž=2(βˆ’6βˆ’4𝑖)βˆ’(βˆ’2βˆ’7𝑖)=βˆ’12βˆ’8𝑖+2+7𝑖=βˆ’10βˆ’π‘–.

Therefore, we can write the equation of the locus of 𝑧 as |𝑧+2+7𝑖|=|π‘§βˆ’(βˆ’10βˆ’π‘–)|.

We will now turn our attention to loci of complex numbers which satisfy equations of the form |π‘§βˆ’π‘§|=π‘˜|π‘§βˆ’π‘§|, where π‘˜>0 and π‘˜β‰ 1. This represents the set of points whose distance from two fixed points is in a constant ratio. One of the ways we could discover what the locus of a point 𝑧 satisfying this equation looks like is to simply manipulate the algebra to find a Cartesian equation for the locus. Doing this in general could be a little laborious and messy, but in the end we would find that we end up with the equation of a circle. We will not do the general derivation here; instead, the next example will look at a specific case of this.

However, we will use geometric reasoning to see that points satisfying this property form a circle.

Consider a circle with center 𝐢 and radius π‘Ÿ. A fixed point 𝐴 lies outside the circle. Another fixed point 𝐡 lies on the line segment between 𝐢 and 𝐴 such that 𝐢𝐴×𝐢𝐡=π‘ŸοŠ¨. Let 𝐷 be a point which lies somewhere on the circle.

Since 𝐷 lies on the circle, 𝐢𝐷=π‘Ÿ. Hence, 𝐢𝐴×𝐢𝐡=𝐢𝐷, which we can rewrite as 𝐢𝐴𝐢𝐷=𝐢𝐷𝐢𝐡.

Using the side-angle-side similarity test, we see that triangles 𝐴𝐢𝐷 and 𝐷𝐢𝐡 are similar. Therefore, 𝐴𝐷𝐷𝐡=𝐢𝐴𝐢𝐷.

However, since point 𝐴 is fixed, 𝐢𝐴 is constant. Furthermore, 𝐢𝐷=π‘Ÿ. Therefore, 𝐢𝐴𝐢𝐷 is a fixed constant for all points on the circle. Therefore, for some fixed π‘˜, we have 𝐴𝐷𝐷𝐡=π‘˜.

Hence, 𝐴𝐷=π‘˜π·π΅ for some fixed π‘˜. This shows us that the locus of a point 𝐷 which satisfies this property is a circle.

Example 6: Describing Loci

A complex number 𝑧 satisfies |𝑧+1βˆ’13𝑖|=3|π‘§βˆ’7βˆ’5𝑖|. Find the Cartesian equation of the locus of 𝑧 and describe it geometrically.

Answer

To find the Cartesian equation of the locus of 𝑧, we substitute 𝑧=π‘₯+𝑦𝑖 into the equation and then we can rearrange it to put it into standard form. Starting with the substitution, we have |π‘₯+𝑦𝑖+1βˆ’13𝑖|=3|π‘₯+π‘¦π‘–βˆ’7βˆ’5𝑖|.

Gathering real and imaginary parts, we have |(π‘₯+1)+(π‘¦βˆ’13)𝑖|=3|(π‘₯βˆ’7)+(π‘¦βˆ’5)𝑖|.

Squaring both sides of the equation yields |(π‘₯+1)+(π‘¦βˆ’13)𝑖|=3|(π‘₯βˆ’7)+(π‘¦βˆ’5)𝑖|.

We can now apply the definition of the modulus to rewrite this as (π‘₯+1)+(π‘¦βˆ’13)=9ο€Ί(π‘₯βˆ’7)+(π‘¦βˆ’5).

Expanding the parentheses, we get π‘₯+2π‘₯+1+π‘¦βˆ’26𝑦+169=9ο€Ήπ‘₯βˆ’14π‘₯+49+π‘¦βˆ’10𝑦+25.=9π‘₯βˆ’126π‘₯+441+9π‘¦βˆ’90𝑦+225.

We can now gather together all of the like terms on the same side of the equation as follows: 8π‘₯βˆ’128π‘₯+8π‘¦βˆ’64𝑦+496=0.

Dividing the whole equation through by 8 gives π‘₯βˆ’16π‘₯+π‘¦βˆ’8𝑦+62=0.

To put this equation into standard form, we can complete the square in both π‘₯ and 𝑦 as follows: (π‘₯βˆ’8)βˆ’64+(π‘¦βˆ’4)βˆ’16+62=0.

Finally, gathering our constants on the right-hand side, we have (π‘₯βˆ’8)+(π‘¦βˆ’4)=18.

Hence, the locus of the point 𝑧 is a circle with center (8,4) and radius 3√2.

We will finish by looking at one final example.

Example 7: Describing Loci

A complex number 𝑧 satisfies |𝑧|+|π‘§βˆ’5βˆ’3𝑖|=8. Describe the locus of 𝑧.

Answer

To understand what this locus represents, we need to consider its geometric meaning. The equation tells us that 𝑧 satisfies the property that the sum of its distance from the origin and 5+3𝑖 is constant and, in particular, equal to 8. At this point, we should consider which geometric shape has this property. In fact, this is the defining feature of an ellipse, that the sum of the distances to the two focal points is constant for every point on the curve. Moreover, from this definition, we conclude that the focal points of the locus are the origin and 5+3𝑖. Finally, what does the value 8 represent? It is the length of the major axis. Hence, the complete description of the locus is that it is an ellipse with foci at the origin and 5+3𝑖 with a major axis of 8.

Key Points

  1. Using our understanding of geometry and the geometry of the complex plane, we can interpret the loci of points which satisfy certain equations.
  2. The locus of a point 𝑧 which satisfies |π‘§βˆ’π‘§|=π‘ŸοŠ§ is a circle of radius π‘Ÿ.
  3. The locus of a point 𝑧 which satisfies |π‘§βˆ’π‘§|=|π‘§βˆ’π‘§| is the perpendicular bisector of the line segment joining π‘§οŠ§ to π‘§οŠ¨.
  4. The locus of a point 𝑧 which satisfies |π‘§βˆ’π‘§|=π‘˜|π‘§βˆ’π‘§|, where π‘˜>0,π‘˜β‰ 1, is a circle.
  5. The locus of a point 𝑧 which satisfies |π‘§βˆ’π‘§|+|π‘§βˆ’π‘§|=π‘ŽοŠ§οŠ¨, where |π‘§βˆ’π‘§|<π‘ŽοŠ§οŠ¨, is an ellipse with foci π‘§οŠ§ and π‘§οŠ¨ with a major axis of length 𝑏.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.