# 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 is the set of all the points which satisfy a particular condition.

Recall that the modulus represents the distance of a point from the origin. For example, given the point , we can calculate the modulus as follows:

Therefore, we know that the point 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 “what is the locus of a point of fixed modulus, say ?” 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 lies at a distance of from and a distance of from . Does the point lie on the circle centered at the origin that passes through and ?

We would like to know whether the point lies on the circle centered at the origin which passes through and . To do this, we could try to calculate the possible positions of and then find its modulus. However, an easier approach would be to try to use the properties of circles to assess whether this lies on the circle. Firstly, we notice that since , they are diametrically opposite. Hence, the line segment from to forms a diameter of the circle. Therefore, will lie on the circle if the triangle formed by , , and 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 , , and satisfy , where 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 and . Recall that we can find the distance between two complex numbers by evaluating the modulus of their difference. Hence, the distance between and is given by

Substituting in the values of and , we have

Using the definition of the modulus, we have

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 and . Hence, we need to verify whether is equal to the sum of the squares of the other two sides:

Since the side lengths satisfy the Pythagorean theorem, we can conclude that the triangle is indeed a right triangle and hence that does lie on the circle passing through and .

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 such that where is constant. Given that is constant, this is the locus of the point which is a fixed distance of from . Hence, this represents a circle centered at of radius .

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 centered at , we can first transform the point to the origin. This translation can be represented by subtracting . Once we have made this translation, a circle of radius is the locus of a point with constant modulus . Hence, the equation representing a circle of radius centered at is given by

### The Loci of Points Forming Circles Centered at 𝑧₁ of Radius 𝑟

For a given constant , the locus of a point , which satisfies , is a circle centered at of radius .

### Example 2: Working with Loci

A complex number satisfies .

1. Describe the locus of and give its Cartesian equation.
2. Find the range of the argument of in the range , correct to three significant figures.
3. Find the range of the modulus of .

Part 1

Since satisfies , its locus represents the points of a constant distance of 2 from the complex number . Hence, it is a circle of radius 2 centered at .

To find the Cartesian equation, we can substitute into the equation and then manipulate it into a standard form as we will demonstrate. Setting , we have

Gathering the real and imaginary parts, we have

Now we can use the definition of the modulus to rewrite this as

Squaring both sides of the equation yields which is a Cartesian equation for a circle of radius 2 centered at .

Part 2

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

We can clearly see that the minimum value of the argument is . We denote the maximum value by . Using the properties of circles, we can see that the triangles and are similar and, hence, the angles at are equal. Therefore, we can find using triangle as follows:

Hence,

Therefore, the range of the argument of is . Evaluating these values, we can write that the range of , correct to three significant figures, is .

Part 3

Finding the range of the modulus is somewhat easier. The maximum and minimum values will be on the half-line extending from to and beyond. We already know the radius of the circle . Hence, minimum and maximum values of will be given by and respectively. Since is the center of the circle represented by , we have that . Therefore, the range of is given by .

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 in the complex plane. Given that is the centre of the circle, write an equation for the locus in the form , where and are constants to be found.

The locus of the point is a circle to write its equation in the form , we need to find its center, represented by point , and its radius, represented by the real number . From the figure, we can see that the locus passes through three points: , , and 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,

Simplifying, we have

To find the center of the circle, we can use the fact that the midpoint of two complex numbers and is given by . Therefore,

Hence, the given figure is the locus of the point satisfying .

The next type of loci we will look at is ones defined by equations of the form , where are constants. The left-hand side represents the distance of from the fixed point , and likewise the right-hand side is the distance of the point from . Therefore, the equation represents the locus of a point which is equidistant from the two points and . Therefore, the locus is the perpendicular bisector of the line segment joining to as shown in the figure.

### The Locus of a Point Equidistant from Two Points

For two given constants , the locus of a point which satisfies is a the perpendicular bisector of the line segment joining to .

### Example 4: Equations of a Locus

A complex number satisfies .

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

Part 1

It is often helpful to put the equation into the standard form to ensure we correctly identify and . Doing this yields

Therefore, the locus of is the perpendicular bisector of the line segment between and .

To find its Cartesian equation, we first calculate the slope of the line segment joining to . From this, we will be able to find the slope of the line representing the locus of .

Using the formula for the slope, we find that

We now need to find the slope representing the locus of which we denote . Since this is perpendicular to this line segment joining to , we can use the property that the product of the slopes of perpendicular lines is . Hence, we have

Therefore, the slope is given by

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 and , the line will pass through the midpoint of the line segment joining to . Recall that we can find the midpoint of two complex numbers by simply taking their average. Hence, the midpoint of the line segment is given by

Simplifying, we get

We can now use the point–slope form of a line, to write the Cartesian equation of the locus of . Substituting in the coordinates of the points representing and the slope of the line, we have

Rearranging, we find that

Part 2

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

This means that we can write a Cartesian equation for the line through the origin and as

Therefore, lies at the point of interaction of this line with the line representing the locus of :

By equating the values of , we can solve for the -coordinate of as follows:

Adding to both sides of the equation yields

Hence,

Substituting this into one of the equations of the lines gives us the value of as follows:

Therefore, the minimum value of is given by

Simplifying gives

### Example 5: Finding the Equations of Loci

The figure shows a locus of a point in the complex plane. Write an equation for the locus in the form , where is a constant to be found.

Firstly, we consider the geometric interpretation of the equation . This equation represents the perpendicular bisector of the line segment from the points represented by and . We can represent this in a diagram as follows.

We will denote the points represented by the complex numbers , , and as , , and . The point represents the intersection of the line segment joining and which we will denote and the line representing the locus of which we will denote . We can therefore find the point by finding the equations of these two lines. Starting with , since it passes through the points and , we can find its slope as follows:

Using the slope–intercept form, we can write the equation of the line as

We now consider the equation of . Since it is perpendicular to , the product of their slopes will be negative one. Hence,

Since this line passes through the point , we can now use the slope–point form, to write its equation as

Subtracting 7 from both sides gives

By equating the equations for and , we can now find the -coordinate of as follows:

Adding and subtracting 4 from both sides of the equation yields

Hence, . Substituting this back into the equation for gives which simplifies to . Therefore, expressed as a complex number the point is .

Finally, to find the value of , we can use the interpretation of complex numbers as vectors. The complex number that represents the vector from to , can be written as . Hence, the complex number which represents the vector from to is given by . Adding this vector to will give us the complex number . Hence,

Substituting in the complex numbers and gives

Therefore, we can write the equation of the locus of as .

We will now turn our attention to loci of complex numbers which satisfy equations of the form , where and . 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 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 and radius . A fixed point lies outside the circle. Another fixed point lies on the line segment between and such that . Let be a point which lies somewhere on the circle.

Since lies on the circle, . Hence, , which we can rewrite as

Using the side-angle-side similarity test, we see that triangles and are similar. Therefore,

However, since point is fixed, is constant. Furthermore, . Therefore, is a fixed constant for all points on the circle. Therefore, for some fixed , we have

Hence, for some fixed . This shows us that the locus of a point which satisfies this property is a circle.

### Example 6: Describing Loci

A complex number satisfies . Find the Cartesian equation of the locus of and describe it geometrically.

To find the Cartesian equation of the locus of , we substitute into the equation and then we can rearrange it to put it into standard form. Starting with the substitution, we have

Gathering real and imaginary parts, we have

Squaring both sides of the equation yields

We can now apply the definition of the modulus to rewrite this as

Expanding the parentheses, we get

We can now gather together all of the like terms on the same side of the equation as follows:

Dividing the whole equation through by 8 gives

To put this equation into standard form, we can complete the square in both and as follows:

Finally, gathering our constants on the right-hand side, we have

Hence, the locus of the point is a circle with center and radius .

We will finish by looking at one final example.

### Example 7: Describing Loci

A complex number satisfies . Describe the locus of .