# Lesson Video: Loci in the Complex Plane Defined Using the Modulus Mathematics

In this video, we will learn how to draw and interpret loci in the complex plane expressed in terms of the modulus.

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### Video Transcript

In this video, we’re going to learn how to draw and interpret loci in the complex plane expressed in terms of the modulus. We’ll begin by looking at geometry in the complex plane before considering loci and the equations of a number of different loci. These will include circular loci, perpendicular bisectors, and elliptical loci.

We begin by recalling some definitions. The locus of a point said in the complex plane is the set of all points said, which satisfy a particular condition. We also know that the modulus of a complex number is the distance of the point representing that complex number plotted on the Argand plane from the origin. For a complex number in algebraic form, 𝑧 equals 𝑎 plus 𝑏𝑖, its modulus is the square root of 𝑎 squared plus 𝑏 squared. In our first example, we’ll simply consider the geometry of the complex plane.

A complex number 𝑤 lies at a distance of five root two from 𝑧 one, which is nine over two plus seven over two 𝑖, and at a distance of four root five from 𝑧 two, which is negative nine over two minus seven over two 𝑖. Does the point 𝑤 lie on the circle centred at the origin that passes through 𝑧 one and 𝑧 two?

To answer this question, we’ll begin by considering the properties of circles. The complex numbers 𝑧 one and 𝑧 two can be plotted on an Argand diagram as shown. 𝑧 one is represented by the point whose cartesian coordinates are nine over two and seven over two. And similarly, 𝑧 two has cartesian coordinates negative nine over two and negative seven over two.

Notice how 𝑧 one is equal to negative 𝑧 two. So they are diametrically opposite. And the line segment that joins 𝑧 one to 𝑧 two must form the diameter of the circle. We can use circle theorems to deduce that this means that 𝑤 will lie on the circle if the triangle formed by 𝑧 one, 𝑧 two, and 𝑤 is a right-angled triangle.

To check whether this is a right-angled triangle, we can use the Pythagorean theorem to check whether the lengths of the sides given as 𝑎, 𝑏, and 𝑐 satisfy the formula 𝑎 squared plus 𝑏 squared equals 𝑐 squared, where 𝑐 is of course the longest side, the hypotenuse, in that triangle.

We already know that 𝑤 is five root two units away from 𝑧 one and that it’s four root five units away from 𝑧 two. So we’ll need to find the length of the third side of the triangle. That’s the length of the side joining 𝑧 one and 𝑧 two. The distance of the length of this side will be the modulus of the difference between these two complex numbers. That’s the modulus of 𝑧 one minus 𝑧 two.

To find their difference, we simply find the difference between their real parts and their imaginary parts. The real parts, that’s nine over two minus negative nine over two. And for the imaginary parts, that’s seven over two minus negative seven over two. And this means that the difference between 𝑧 one and 𝑧 two is nine plus seven 𝑖.

Remember, we’re looking to find the modulus of the difference between these two numbers. So that’s the square root of nine squared plus seven squared, which is the square root of 130.

Now that we know the lengths of these three sides, we can check whether they satisfy the Pythagorean theorem. Root 130 is longer than five root two and four root five. So we’re going to find the sum of the squares of the two shorter lengths. Four root five squared is 50, and five root two squared is 80. 50 plus 80 is 130, which is indeed equal to the square of root 130.

We’ve seen that these three sides satisfy the Pythagorean theorem and therefore form the sides of a right-angled triangle. This in turn means that the line joining 𝑧 one and 𝑧 two form the diameter of a circle for which 𝑤 lies on the circumference as required.

Let’s look to generalise this idea. We saw that the distance between two points is given by the modulus of their difference. We can therefore say that, for a given constant complex number 𝑧 one, the locus of a point 𝑧, which satisfies the equation the modulus of 𝑧 minus 𝑧 one is equal to 𝑟, is a circle centred at 𝑧 one of radius 𝑟. Let’s now consider an example of the application of this definition.

A complex number 𝑧 satisfies the equation the modulus of 𝑧 minus two plus three 𝑖 is equal to two. 1) Describe the locus of 𝑧 and give it a cartesian equation. 2) Find the range of the argument of 𝑧 in the interval negative 𝜋 to 𝜋. 3) Find the range of the modulus of 𝑧.

Remember, for a constant complex number 𝑧 one, the locus of a point 𝑧, which satisfies the equation given, is a circle centred at 𝑧 one with a radius 𝑟. To answer part one then, we’ll rewrite 𝑧 minus two plus three 𝑖 slightly by factoring negative one. And we get 𝑧 minus two minus three 𝑖. This means that since our complex number 𝑧 satisfies this equation, its locus is a circle with a radius of two, whose centre is at two minus three 𝑖.

And there are two ways we can give this a cartesian equation. We could substitute 𝑧 equals 𝑥 plus 𝑦𝑖 into the given equation. Alternatively, we recall the cartesian equation for a circle centred at 𝑎𝑏 with a radius of 𝑟. It’s 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared equals 𝑟 squared.

Taking 𝑥 to be the real part and 𝑦 to be the imaginary part, we know that, for our circle, the radius is two, 𝑎 is equal to two, and 𝑏 is equal to negative three. Substituting these into the formula, and we get 𝑥 minus two all squared plus 𝑦 minus negative three all squared equals two squared. So the cartesian equation simplifies to 𝑥 minus two squared plus 𝑦 plus three squared equals four.

And now we consider part two. Find the range of the argument of 𝑧 in the interval negative 𝜋 to 𝜋.

We’ll begin by drawing the locus given on an Argand diagram. Remember, it’s a circle whose centre lies at two, negative three. And its radius is two units. This means that the imaginary axis must be a tangent to this circle. It’s quite clear then that the smallest possible value for the argument must be negative 𝜋 by two radians. But what about its maximum value?

Well, let’s call that 𝜃 minus 𝜋 by two. And we’ll add another tangent to the circle. We’ll call this at point 𝑏. We know that triangles 𝑂𝐵𝐶 and 𝐴𝑂𝐶 are congruent. They are right-angled triangles who share a hypotenuse of equal length. They also both have the radius of the circle as one of their sides. So they must be congruent. This means these acute angles at 𝑂 must be equal. We’ll call them 𝜃 divided by two.

Using right angle trig, we can see that tan of 𝜃 divided by two must be equal to 𝐴𝐶 divided by 𝐴𝑂. 𝐴𝐶 is two units, and 𝐴𝑂 is three units. So 𝜃 divided by two must be equal to arc tan of two-thirds. And we can therefore say that 𝜃 is equal to two times arctan of two-thirds. And the maximum value is therefore two arctan of two-thirds minus 𝜋 by two. And we have the range for the argument.

So let’s look at part three. Find the range of the modulus of 𝑧.

We know that the minimum value of 𝑧 will lie at this point 𝐷. And the maximum will lie at the point 𝐸. And that’s because the modulus is the distance between the point representing the complex number on the Argand diagram and the origin. And we actually know the radius of the circle. So we can define the minimum as the length of 𝑂𝐶 minus the radius and the maximum as the length of 𝑂𝐶 plus the radius.

Now of course we saw that the radius is two. So the minimum is 𝑂𝐶 minus two, and the maximum is the length of 𝑂𝐶 plus two. And we can use the distance formula or the definition of the modulus to find the length of 𝑂𝐶. 𝐶 is at the point two, negative three. So the length of 𝑂𝐶 is the square root of two squared plus negative three squared, which is root 13. And we can see that the range of the modulus of 𝑧 is values between root 13 minus two and root 13 plus two. We can also reverse this process and apply standard coordinate geometry to allow us to describe a given locus as an equation in terms of the modulus.

Let’s see what that might look like.

The figure shows a locus of a point 𝑧 in the complex plane. Write an equation for the locus in the form the modulus of 𝑧 minus 𝑎 equals 𝑏, where 𝑎 is a complex number and 𝑏 is greater than zero, and they’re constants to be found.

Remember, the locus of a point 𝑧, which satisfies the equation the modulus of 𝑧 one is 𝑧 one equals 𝑟, is a circle centred at 𝑧 one of radius 𝑟. We therefore need to find the centre and radius of the circle represented on our diagram.

We see that the circumference of the circle passes through three points. These are the points that represent the complex numbers zero, four 𝑖, and negative 10. We see that the vertex at 𝑧 one is right. And then we therefore know that the line joining 𝑧 one and 𝑧 two passes through the centre of the circle. It’s the diameter of the circle.

We can use the Pythagorean theorem to calculate its length. And thus we’ll be able to find the radius. Denoting the diameter as 𝑑, we see that 𝑑 is equal to the square root of 10 squared plus four squared. And that’s equal to two root 29. Its radius is half of this. So the radius of our circle is root 29 units.

We also know that the centre of the circle must lie at the midpoint of the diameter. We could apply standard coordinate geometry here. Or we could recall the fact that the midpoint of two complex numbers is half of their sum. And we see that the centre of the circle lies at the point which represents a complex number a half of negative 10 plus four 𝑖. That’s negative five plus two 𝑖. So we have a circle whose radius is root 29 and whose centre lies at negative five plus two 𝑖. This means that the equation of our circle and therefore the equation of the locus given is the modulus of 𝑧 minus negative five plus two 𝑖 equals root 29.

The next example, we use the fact that an equation given by the modulus of 𝑧 minus 𝑧 one equals the modulus of 𝑧 minus 𝑧 two represents a perpendicular bisector of the line segment which joins the points 𝑧 one to 𝑧 two. For example, the modulus of 𝑧 equals the modulus of 𝑧 minus six 𝑖 represents the locus of all points equidistant from the points zero, zero and zero, six. Let’s look at a slightly more complicated example.

A complex number 𝑧 satisfies the modulus of 𝑧 plus one plus 𝑖 equals the modulus of 𝑧 minus two minus six 𝑖. Describe the locus of 𝑧 and give its cartesian equation. We’ll begin by factoring the terms inside each modulus to ensure that this locus looks like the general form.

The modulus of 𝑧 minus 𝑧 one equals the modulus of 𝑧 one minus 𝑧 two. That’s the modulus of 𝑧 minus negative one minus 𝑖 equals the modulus of 𝑧 minus two plus six 𝑖. This means that the locus of 𝑧 is given by all points equidistant from negative one minus 𝑖 and two plus six 𝑖. It’s the perpendicular bisector of the line segment between the two points on the Argand plane.

We can find its cartesian equation as we would the cartesian equation of any line, by first finding its gradient. The gradient of the line segment between these points representing our two complex numbers is six minus negative one over two minus negative one, which is seven-thirds. Since the locus of the points representing 𝑧 form the perpendicular line bisector of this line segment, the gradient is found by using the fact that the product of the gradients of two lines, which are perpendicular, is negative one. So its gradient is negative three-sevenths.

We also know that this line passes through the midpoint of the points representing our two complex numbers. And this midpoint is half the sum of these two complex numbers. It’s a half of negative one minus 𝑖 plus two plus six 𝑖. That’s a half plus five over two 𝑖. Using the formula 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one, with the cartesian coordinates a half, five over two, we get 𝑦 minus five over two equals negative three-sevenths times 𝑥 minus one-half.

And we can rearrange this. And we see that the equation of our line is 𝑦 equals negative three-sevenths 𝑥 plus 19 over seven. For our two final examples, we’ll consider the locus of a point 𝑧 that forms a circle different to our first example and that which forms an ellipse. The first definition we need to know is that the locus of a point 𝑧, which satisfies the equation the modulus of 𝑧 minus 𝑧 one is equal to 𝑘 times the modulus of 𝑧 minus 𝑧 two when 𝑘 is greater than zero and not equal to one, is a circle. We also need to know that the locus of a point 𝑧, which satisfies the equation the modulus of 𝑧 minus 𝑧 one plus the modulus of 𝑧 minus 𝑧 two is equal to 𝑎, where the modulus of 𝑧 one minus 𝑧 two is less than 𝑎, is an ellipse, with foci 𝑧 one and 𝑧 two and a major axis of length 𝑏.

A complex number 𝑧 satisfies the equation the modulus of 𝑧 plus one minus 13𝑖 equals three times the modulus of 𝑧 minus seven minus five 𝑖. Find the cartesian equation of the locus of 𝑧.

We know that the locus of the points 𝑧 which satisfy this equation form a circle. We can find its centre and radius by substituting the general algebraic form of the complex number into this equation. We’ll let 𝑧 be equal to 𝑥 plus 𝑦𝑖. On the left-hand side, we get 𝑥 plus 𝑦𝑖 plus one minus 13𝑖. And on the right-hand side, it becomes 𝑥 plus 𝑦𝑖 minus seven minus five 𝑖. We can gather real and imaginary parts.

And next we consider the definition of the modulus. The modulus of a complex number is the square root of the sum of the squares of the real and imaginary parts. In our case, that’s as shown. We’re going to square both sides of this equation. And next, we distribute inside our parentheses and gather all the terms. We can then divide through by eight. And we have 𝑥 squared minus 16𝑥 plus 𝑦 squared minus eight 𝑦 plus 62 equals zero.

We can then complete the square. And we see that this simplifies to 𝑥 minus eight all squared plus 𝑦 minus four all squared equals 18. We found the cartesian equation of the locus of 𝑧. And in fact, we can describe this locus as a circle, which we said earlier. However, we now know it has a centre at the point whose cartesian coordinates are eight, four. And its radius is root 18, which simplifies to three root two units.

A complex number satisfies the modulus of 𝑧 plus the modulus of 𝑧 minus five minus three 𝑖 equals eight. Describe the locus of 𝑧.

Remember, the locus of a point 𝑧 which satisfies the modulus of 𝑧 minus 𝑧 one plus the modulus of 𝑧 minus 𝑧 two equals 𝑎, where the modulus of 𝑧 one minus 𝑧 two is less than 𝑎, is an ellipse, with a foci at 𝑧 one and 𝑧 two and a major axis of length 𝑏. We can rewrite our equation slightly to be the modulus of 𝑧 minus zero plus the modulus of 𝑧 minus five plus three 𝑖 equals eight. This means the locus of 𝑧 is an ellipse. It has foci at the origin and five plus three 𝑖. And it has a major axis of length eight units.

In this lesson, we’ve seen that we can use our understanding of geometry and the geometry of the complex plane to interpret the loci of points which satisfy certain equations. We’ve seen that the locus of a point 𝑧, which satisfies this equation, is a circle of radius 𝑟. The locus of the points which satisfy these equations are the perpendicular bisector of the line segment between 𝑧 one and 𝑧 two. We’ve also seen the alternative form for the locus of the points which are a circle, as well as the locus of points which are an ellipse.