Video: Loci in the Complex Plane Defined Using the Argument

Video Transcript

In this video, we’ll learn how to draw and interpret loci in the complex plane, expressed in terms of the argument. Just as we can use the modulus to define loci in the complex plane, by considering the geometry of this plane, we can also use the arguments of a complex number to interpret the loci of points which satisfy certain criteria. We’ll consider the loci of half planes, major arcs, semicircles, and minor arcs, and the Cartesian equations which correspond to these in this video.

Remember, for a complex number represented on an Argand diagram which is joined by a line segment or a half line to the origin, the argument is the angle this line segment makes with the positive real axis. And it’s always measured in a counterclockwise direction. To calculate the argument, we first consider which quadrant the point representing the complex number will lie in. For a complex number of the form 𝑧 equals 𝑎 plus 𝑏𝑖, its argument is given as the arctan of 𝑏 divided by 𝑎, for complex numbers which would be represented in the first and fourth quadrant. For complex numbers which will be represented in the second quadrant, the argument is the arctan of 𝑏 divided by 𝑎 plus 𝜋. And if we’re looking at a complex number that’s plotted in the third quadrant, its argument is arctan of 𝑏 divided by 𝑎 minus 𝜋.

Now, rather than asking what the argument of a complex number is, we could alternatively ask what is the locus of a point of a fixed argument, say the argument of 𝑧 is equal to 𝜋 by three radians? This represents the set of complex numbers that lie on the ray or the half line which makes an angle of 𝜋 by three with the 𝑥-axis in a counterclockwise direction. The locus of 𝑧 is therefore this half line. Remember, though, the argument is not defined when 𝑧 is equal to zero. So the locus cannot include the origin.

We can generalize this for any half line in the complex plane, by considering a transformation by subtracting a fixed complex number, 𝑧 one. We can say that the locus of the point 𝑧 such that the argument of 𝑧 minus 𝑧 one is equal to 𝜃 is a half line from but not including the point at 𝑧 one. This half line makes an angle of 𝜃 with the horizontal half line extending from 𝑧 one in the positive 𝑥-direction. And it’s measured in a counterclockwise direction. Let’s have a look at an example of this.

Sketch the locus of 𝑧 when the argument of 𝑧 plus two plus 𝑖 is equal to 𝜋 by four.

Remember, the locus of the point 𝑧 when the argument of 𝑧 minus 𝑧 one is equal to 𝜃 is a half line from but not including 𝑧 one. This half line makes an angle of 𝜃 to the horizontal half line in the positive 𝑥-direction. And it’s measured in a counterclockwise direction. We’ll begin then by writing the argument of 𝑧 plus two plus 𝑖 in the form the argument of 𝑧 minus 𝑧 one, to ensure that we can correctly identify our value of 𝑧 one. We factor negative one, and we get the argument of 𝑧 minus negative two minus 𝑖. And this means we can rewrite our equation as the argument of 𝑧 minus negative two minus 𝑖 equals 𝜋 by four.

We can now see that 𝑧 one is equal to negative two minus 𝑖. This is the endpoint of the ray or half line. And we do need to recall that the half line doesn’t actually include this point. On an Argand diagram, 𝑧 one can be plotted by the point whose Cartesian coordinates are negative two, negative one as shown. And we’ve added this empty circle to show that we don’t want to include this point in our locus.

Next, we use the argument. The argument of 𝑧 plus two plus 𝑖 is 𝜋 by four radians. This means the locus of 𝑧 is the set of points that make an angle of 𝜋 by four radians in a counterclockwise direction from the horizontal. 𝜋 by four radians is equivalent to 45 degrees. So we add a line, as shown, at this angle. And this means that the locus of 𝑧 when the argument of 𝑧 plus two plus 𝑖 is equal to 𝜋 by four is as shown. We can also reverse this process to form the equation given a diagram of the locus of 𝑧.

In our next example, we’ll look at how the locus can be expressed as a Cartesian equation.

Find the Cartesian equation of the locus of 𝑤 such that the argument of 𝑤 plus three plus 𝑖 is equal to 𝜋 by three.

Remember, the locus in this form is a half line. We’re looking to find the Cartesian equation of this half line. So a sensible starting point is to find the gradient of this line. We can find the gradient of this line by considering the argument, which is 𝜋 by three radians. Now, the formula for gradient is rise over run. That’s the same as opposite over adjacent. And, of course, that’s equal to the tangent function. We can say then that the gradient of our line is equal to tan of 𝜋 by three which is the square root of three.

Our next job is to find the point which this line must pass through. We use the definition of the locus to rewrite our equation. We factor negative one. And we can see that this is the same as the argument of 𝑤 minus negative three minus 𝑖 equals 𝜋 by three. And we can see then that our line begins at the point representing the complex number negative three minus 𝑖. This will have Cartesian coordinates negative three, negative one. Let’s substitute these values into the formula for a straight line, 𝑦 minus 𝑦 one equals 𝑚 multiplied by 𝑥 minus 𝑥 one.

When we do, we see that 𝑦 minus negative one equals root three times 𝑥 minus negative three. We distribute the parentheses and simplify as far as possible. And we can see that the Cartesian equation of the line is 𝑦 equals root three 𝑥 plus three root three minus one. Remember, though, this is a half line. And it doesn’t actually include the point at negative three, negative one. So we’re going to need to add a restriction on 𝑥 or 𝑦. We can say that 𝑥 must be greater than negative three. And we found the Cartesian equation of the locus of 𝑤, given that the argument of 𝑤 plus three plus 𝑖 is 𝜋 by three.

The next locus we are interested in is that of a circle. We can say that the locus of the point 𝑧 such that the argument of 𝑧 minus 𝑧 one over 𝑧 minus 𝑧 two is equal to 𝜃 is the arc of a circle which subtends an angle of 𝜃 between the points represented by 𝑧 one and 𝑧 two, as shown in the diagram. If 𝜃 is less than 𝜋 by two radians, the locus is a major arc. If 𝜃 is equal to 𝜋 by two, the locus is a semicircle. And if 𝜃 is greater than 𝜋 by two radians, the locus is a minor arc. Now, remember, the endpoints are not part of the locus. So we include open dots representing these points, as shown. Let’s look at an example that uses this idea.

The figure shows a locus of a point 𝑧 in the complex plane. Write an equation for the locus in the form the argument of 𝑧 minus 𝑎 over 𝑧 minus 𝑏 equals 𝜃, where 𝑎 and 𝑏, which are complex numbers, and 𝜃, which is greater than zero and less than or equal to 𝜋, are constants to be found.

Remember, the locus of a point 𝑧 in this form is the arc of a circle which subtends an angle of 𝜃 between the points represented by 𝑧 one and 𝑧 two. We have three conditions on 𝜃. If it’s less than 𝜋 by two, the locus is a major arc. If it’s equal to 𝜋 by two, it’s a semicircle. And if it’s greater than 𝜋 by two, the locus is a minor arc. And, remember, the endpoints are not part of this locus. We can see by looking at the diagram that the locus of our 𝑧 is the major arc of a circle. And this makes sense because 𝜃 is equal to 𝜋 by five radians.

The endpoints of our locus lie at 𝐴 and 𝐵 whose Cartesian coordinates are four, negative three and negative three, one, respectively. These represent the complex numbers four minus three 𝑖 and negative three plus 𝑖. And, remember, this locus is traced in a counterclockwise direction. Since the starting point is that represented in the complex number four minus three 𝑖, we can say that the equation of our locus is the argument of 𝑧 minus four minus three 𝑖 over 𝑧 minus negative three plus 𝑖 equals 𝜋 by five. We were actually told to find the value of the constants 𝑎, 𝑏, and 𝜃. 𝑎 is four minus three 𝑖, 𝑏 is negative three plus 𝑖, and 𝜃 is 𝜋 by five.

In our next example, we’ll practice sketching a locus in this form.

The point 𝑧 satisfies the equation the argument of 𝑧 minus six over 𝑧 minus six 𝑖 equals 𝜋 by four. Plot the locus of 𝑧 on an Argand diagram.

The locus of 𝑧 is the arc of a circle which subtends an angle of 𝜋 by four radians between the points represented by six and six 𝑖. Remember, these are plotted in a counterclockwise direction from six to six 𝑖. But they don’t actually include these points themselves. These are the points on the Argand plane whose Cartesian coordinates are six, zero and zero, six, respectively. And since 𝜋 by four is less than 𝜋 by two, we know that we have a major arc. So we begin by adding the points six, zero and zero, six on our Argand diagram.

And then we come across a problem. How do we know where the major arc sits? Sure, it’s the major arc of a circle. But without knowing the centre of the circle, we can’t use that information to find the arc. It could actually be either of these two arcs shown. Here, we recall the fact that the locus is drawn in a counterclockwise direction. We need the arc that begins at the point six, zero and ends up at the point zero, six to be a major arc, when drawn in this direction.

This means we have to choose this arc on the right. And, therefore, the locus is as shown. It’s not actually necessary to add the cords shown. But by doing so, we can see that we will get an angle that’s less than 𝜋 by two radians. It is also possible to find the Cartesian equation of loci in this form. Occasionally, you can take a geometric approach. But, in general, an algebraic approach is sensible.

In our final example, we’ll consider one such algebraic approach.

The locus of 𝑧 satisfies the equation the argument of 𝑧 minus three 𝑖 over 𝑧 minus five 𝑖 equals two 𝜋 by three. Plot this locus on an Argand diagram and find its Cartesian equation.

Remember, the locus of the point 𝑧 such that the argument of 𝑧 minus three 𝑖 over 𝑧 minus five 𝑖 equals two 𝜋 by three is the arc of a circle which subtends an angle of two 𝜋 by three radians between the points represented by three 𝑖 and five 𝑖. This time, it will begin at three 𝑖 and travel in a counterclockwise direction. Two 𝜋 by three is greater than 𝜋 by two. So we know that this forms a minor arc. And as always, the endpoints are not part of the locus. The points represented by three 𝑖 and five 𝑖 have Cartesian coordinates zero, three and zero, five, respectively.

So once we’ve plotted these points on an Argand diagram, how do we decide where the minor arc sits? Once again, we don’t know the centre of the circle. So we can’t use that information to find the location of the arc. We do, however, know that the locus is drawn in a counterclockwise direction. We need the arc from zero, three to zero, five to be a minor arc, when drawn in that counterclockwise direction. This means we travel along this arc as show.

Next, we need to find the Cartesian equation of this locus. In some scenarios, we can find this equation by finding the centre and radius of the circle. Here, that’s not so easy. So we’re going to need to substitute 𝑧 equals 𝑥 plus 𝑦𝑖 into our equation. When we do, we get that the argument of 𝑥 plus 𝑦𝑖 minus three 𝑖 over 𝑥 plus 𝑦𝑖 minus five 𝑖 is two 𝜋 by three. Let’s begin by evaluating 𝑥 plus 𝑦𝑖 minus three 𝑖 over 𝑥 plus 𝑦𝑖 minus five 𝑖. To evaluate this problem, we need to multiply both the numerator and the denominator of our fraction by the conjugate of the denominator.

To find the conjugate of a complex number, we change the sign for the imaginary part. So the conjugate of 𝑥 plus 𝑦 minus five 𝑖 is 𝑥 minus 𝑦 minus five 𝑖. We’re going to multiply both the numerator and the denominator of this fraction by this number. On the numerator, we end up with 𝑥 squared minus 𝑥 𝑦 minus five 𝑖 plus 𝑥 𝑦 minus three 𝑖 minus 𝑦 minus three times 𝑦 minus five times 𝑖 squared. And on the denominator, we have 𝑥 squared minus 𝑥 times 𝑦 minus five times 𝑖 plus 𝑥 times 𝑦 minus five times 𝑖 minus 𝑦 minus five squared 𝑖 squared. And we can see that negative 𝑥 times 𝑦 minus five times 𝑖 plus 𝑥 times 𝑦 minus five times 𝑖 is zero.

Then using the fact that 𝑖 squared is equal to negative one and distributing our parentheses, we have the expression shown. Next, we collect the real and imaginary parts. And now we can find the argument of 𝑥 plus 𝑦𝑖 minus three 𝑖 over 𝑥 plus 𝑦𝑖 minus five 𝑖. If we take 𝑎 to be the real part of our complex number and 𝑏 to be the imaginary part, that is two 𝑥 over 𝑥 squared plus 𝑦 squared minus 10𝑥 plus 25, we can say that 𝑏 divided by 𝑎, the imaginary part divided by the real part, must be equal to tan of two 𝜋 by three. Now, normally, we would be worrying about which quadrant the complex number lies in. But since time is periodic with a period of 𝜋 radians, adding or subtracting multiples of 𝜋 to our value of 𝜃 has no effect on the value of tan 𝜃. Let’s clear some space for the next step.

We can say that two 𝑥 divided by 𝑥 squared plus 𝑦 squared minus eight 𝑦 plus 15 is equal to tan of two 𝜋 by three, which is equal to negative root three. We multiply both sides of this equation by 𝑥 squared plus 𝑦 squared minus eight 𝑦 plus 15. And then we can simplify a little by multiplying through by negative root three. And then, we add two root three 𝑥 to both sides of this equation. We now need to complete the square for both 𝑥 and 𝑦.

Remember, we’re trying to find the equation of the circle. That’s 𝑥 plus root three all squared minus three plus 𝑦 minus four all squared minus 16 plus that 15. Negative three minus 16 plus 15 is negative four. So we add four to both sides of this equation. And we get 𝑥 plus root three all squared plus 𝑦 minus four all squared equals four. And we can see that the centre of our circle now lies at the point with Cartesian coordinates negative root three, four. We do, of course, need to add a restriction on 𝑥 and 𝑦 to ensure that the points three 𝑖 and five 𝑖 don’t actually lie on the locus itself. This restriction is that 𝑥 is greater than zero. So the Cartesian equation of our locus is four equals 𝑥 plus root three all squared plus 𝑦 minus four all squared, only when 𝑥 is greater than zero.

In this video, we’ve seen the we can use the arguments in the same way that we can use the modulus to define loci in the complex plane. We’ve seen that the locus of a point 𝑧 which satisfies the argument of 𝑧 minus 𝑧 one equals 𝜃 is a half line from but not including 𝑧 one. And it makes an angle of 𝜃 to the horizontal half line extending from 𝑧 one in the positive 𝑥-direction.

We’ve also seen that 𝜃 must be measured in a counterclockwise direction. We saw how the locus of a point 𝑧 which satisfies the equation the argument of 𝑧 minus 𝑧 one over 𝑧 minus 𝑧 two equals 𝜃 is an arc. When 𝜃 is less than 𝜋 by two, it’s a major arc. When it’s equal to 𝜋 by two, it’s a semicircle. And when it’s greater than 𝜋 by two, the locus is a minor arc. And we saw that the endpoints cannot be part of this locus. And the locus is measured in a counterclockwise direction.