Explainer: Loci in the Complex Plane Using the Argument

In this explainer, we will learn how to find the loci of a complex equation in the complex plane defined in terms of the argument.

Recall that the argument of a complex number represents the angle that the half-line from the origin to the complex number makes with the positive real axis (๐‘ฅ-axis), measured in a counterclockwise direction (often referred to as the positive angle or in the positive direction).

For example, given the point ๐‘ค=โˆ’1+๐‘–โˆš3, to calculate the argument, we need to consider which of the quadrants of the complex plane the number lies in. In this case, we have a number in the second quadrant. This means that we need to add ๐œ‹ to the result we get from the inverse tangent. Hence, argarctan(๐‘ค)=๏€ปโˆ’โˆš3๏‡+๐œ‹=โˆ’๐œ‹3+๐œ‹=2๐œ‹3.

Geometrically, this tells us that ๐‘ค lies on a ray (or half line) extending radially outwards from the origin making an angle of 2๐œ‹3 with the ๐‘ฅ-axis in the counterclockwise direction.

Rather than asking what the argument of a particular complex number is, we could ask the following question: what is the locus of a point ๐‘ง of a fixed argument, say, arg(๐‘ง)=2๐œ‹3. This represents all the complex numbers which lie on the ray which makes an angle of 2๐œ‹3 with the ๐‘ฅ-axis in the counterclockwise direction. Hence, we can simply state that the locus of ๐‘ง such that arg(๐‘ง)=2๐œ‹3 is this ray. However, it is important to remember that since the argument is not defined when ๐‘ง=0, the locus will not include the origin. Note that a ray which does not include its end point is often referred to as an open ray or as an open half-line.

We can of course transform the equation arg(๐‘ง)=๐œƒ by subtracting a fixed ๐‘งโˆˆโ„‚๏Šง. This will result in a half-line whose endpoint is at ๐‘ง๏Šง in the plane.

Loci Representing Half-Lines in the Plane

The locus of the point ๐‘ง such that arg(๐‘งโˆ’๐‘ง)=๐œƒ๏Šง is a half-line from, but not including, ๐‘ง๏Šง which makes an angle of ๐œƒ to the horizontal half-line extending from ๐‘ง๏Šง in the positive ๐‘ฅ-direction measured in a counterclockwise direction.

Example 1: Identifying Loci

Which of the following figures is the correct representation of the locus of ๐‘ง which satisfies arg(๐‘งโˆ’2+3๐‘–)=๐œ‹8?

Answer

We begin by writing the equation of the locus in the standard form arg(๐‘งโˆ’๐‘ง)=๐œƒ๏Šง to ensure that we correctly identify the value of ๐‘ง๏Šง. Noting that ๐‘งโˆ’2+3๐‘–=๐‘งโˆ’(2โˆ’3๐‘–), we rewrite the equation as arg(๐‘งโˆ’(2โˆ’3๐‘–))=๐œ‹8.

By writing it in this form, we can easily identify that ๐‘ง=2โˆ’3๐‘–๏Šง. This is the end point of the ray.

Looking at the options, there are three figures in which the endpoint of the ray is at 2โˆ’3๐‘–: (a), (b), and (d). Recall that modulus is measured in the counterclockwise direction. Therefore, given that ๐œ‹8 is positive, the answer is not figure (b). We now have two options to consider: (a) and (d). The difference between these two figures is the whether we have used a solid circle, โ€ข, or a hollow circle, โˆ˜, to represent the end points. Recall that we use the solid circle to indicate that the line includes the endpoint, 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 locus of ๐‘ง will not include the end point and hence the correct representation for the locus is figure (d).

It is important to remember to use the correct notation for the end points when sketching loci. This is particularly true for exam questions where you may lose marks for not correctly highlighting whether or not the end points are included in your loci.

Example 2: Finding the Equation of a Locus

The figure shows a locus of a point ๐‘ง in the complex plane. Write an equation for the locus in the form arg(๐‘งโˆ’๐‘Ž)=๐œƒ, where ๐‘Žโˆˆโ„‚ and โˆ’๐œ‹<๐œƒโ‰ค๐œ‹ are constants to be found.

Answer

Recall that for a locus that satisfies arg(๐‘งโˆ’๐‘Ž)=๐œƒ, ๐‘Ž represents the end point of the ray and ๐œƒ the directed angle it makes to the horizontal half-line extending from ๐‘ง๏Šง in the positive ๐‘ฅ direction measured in a counterclockwise direction.

From the figure, we can see that the endpoint of the ray is at the point with coordinates (โˆ’1,2). Therefore, the complex number, ๐‘Ž, that represents this point will be given by ๐‘Ž=โˆ’1+2๐‘–. We now consider the angle that the ray makes to the horizontal. The angle given in the figure is measured from the positive horizontal from ๐‘ง๏Šง as expected. However, since the angle represents a clockwise rotation, we should take its negative since the argument is defined in terms of angles measured in the counterclockwise direction. Hence, ๐œƒ=โˆ’6๐œ‹7. Therefore, the equation of the locus is given by arg(๐‘งโˆ’(โˆ’1+2๐‘–))=โˆ’6๐œ‹7.

Example 3: Loci

Consider ๐‘ง and ๐‘ค in the complex plane.

  1. Find the Cartesian equation of the locus of ๐‘ง such that arg(๐‘งโˆ’2โˆ’๐‘–)=5๐œ‹6.
  2. Find the Cartesian equation of the locus of ๐‘ค such that arg(๐‘ค+3+๐‘–)=๐œ‹3.
  3. Find the point at which the two loci meet and the angle they meet at.

Answer

We can find the Cartesian equation of the locus algebraically or geometrically. We will use the algebraic method for part 1 and the geometric method for part 2.

Part 1

To find the Cartesian equation, we start by substituting ๐‘ง=๐‘ฅ+๐‘ฆ๐‘– into the equation as follows: arg(๐‘ฅ+๐‘ฆ๐‘–โˆ’2โˆ’๐‘–)=5๐œ‹6.

Gathering real and imaginary parts, we have arg((๐‘ฅโˆ’2)+(๐‘ฆโˆ’1)๐‘–)=5๐œ‹6.

Using the definition of the argument in terms of the tangent function, we can rewrite this as ๐‘ฆโˆ’1๐‘ฅโˆ’2=๏€ผ5๐œ‹6๏ˆ.tan

Notice that we need not worry about which quadrant the complex number is in, since adding or subtracting multiples of ๐œ‹ to ๐œƒ has no effect on the value of tan๐œƒ. Evaluating the tan on the right-hand side of the equation, we have ๐‘ฆโˆ’1๐‘ฅโˆ’2=โˆ’1โˆš3.

Multiplying both sides of the equation by ๐‘ฅโˆ’2 gives ๐‘ฆโˆ’1=โˆ’1โˆš3(๐‘ฅโˆ’2).

Adding one to both sides and simplifying give us ๐‘ฆ=โˆ’โˆš33๐‘ฅ+1+2โˆš33.

However, we are not quite finished; we have given the equation of a line and not of a half-line. Therefore, we need to add a condition to either ๐‘ฅ or ๐‘ฆ to restrict our equation to a half-line. We will add our restriction to ๐‘ฅ. We only want the points to the right of ๐‘ฅ=2. Hence, our condition on ๐‘ฅ is ๐‘ฅ<2. Note that we use less than and not less than or equal to since the half-line is open; that is, it does not include the endpoint. Therefore, the Cartesian equation of the locus of ๐‘ง is given by ๐‘ฆ=โˆ’โˆš33๐‘ฅ+1+2โˆš33,๐‘ฅ<2.where

Part 2

Recall that a locus in this form is a half-line or ray. Therefore, we will start by finding the slope of the line. Since we are given the angle the line makes to the horizontal, to find the slope ๐‘š, we can simply take its tangent. Hence, ๐‘š=๏€ป๐œ‹3๏‡=โˆš3.tan

Now we need a point that the line passes through. Using the complex equation of the locus, we can see that the line will pass through the point (โˆ’3,โˆ’1). Therefore, we can use the slopeโ€“intercept form, ๐‘ฆโˆ’๐‘ฆ=๐‘š(๐‘ฅโˆ’๐‘ฅ),๏Šง๏Šง to write the equation of the line as ๐‘ฆ+1=โˆš3(๐‘ฅ+3).

Since we know that the locus only represents the half-line, we need to add a condition on ๐‘ฅ or ๐‘ฆ. Hence, the Cartesian equation of the locus of ๐‘ค is given by ๐‘ฆ=โˆš3๐‘ฅ+3โˆš3โˆ’1,๐‘ฅ>โˆ’3.where

Part 3

To find the point where the two loci intersect, we can equate the equations we derived in parts 1 and 2 and solve for ๐‘ฅ as follows: โˆš3๐‘ฅ+3โˆš3โˆ’1=โˆ’โˆš33๐‘ฅ+1+2โˆš33.

Notice that this equation is only valid for โˆ’3<๐‘ฅ<2. Therefore, if we find solutions outside of this range, the two loci do not intersect. Gathering the ๐‘ฅ terms on the left-hand side and the constant terms of the right-hand side yields 43โˆš3๐‘ฅ=2โˆ’73โˆš3.

Hence, ๐‘ฅ=6โˆ’7โˆš34โˆš3=โˆš32โˆ’74.

Notice that this solution is in the range โˆ’3<๐‘ฅ<2. Therefore, the loci intersect. To find the corresponding value of ๐‘ฆ, we can substitute this value of ๐‘ฅ back into the equation we got in part 2 as follows: ๐‘ฆ=โˆš3๏€ฟโˆš32โˆ’74๏‹+3โˆš3โˆ’1=32โˆ’74โˆš3+3โˆš3โˆ’1=12+54โˆš3.

Therefore, the two loci intersect at the point ๏€ฟโˆš32โˆ’74,12+54โˆš3๏‹. To find the angle that the two loci cross at, we plot them on an Argand diagram.

Using the property of alternative angles, we can see that angle ๐œƒ is equal to the difference of the two angles. Hence, ๐œƒ=5๐œ‹6โˆ’๐œ‹3=๐œ‹2.

In the previous example, we considered the point of intersection of two loci of the forms arg(๐‘งโˆ’๐‘Ž)=๐›ผ and arg(๐‘คโˆ’๐‘)=๐›ฝ. We saw that the angle that they intersected at was given by ๐œƒ=๐›ผโˆ’๐›ฝ. Rather than fixing the value of ๐›ผ and ๐›ฝ, what would happen if we fixed the value of ๐œƒ=๐›ผโˆ’๐›ฝ and considered the locus of the point ๐‘ง such that arg(๐‘งโˆ’๐‘Ž)=๐›ผ and arg(๐‘งโˆ’๐‘)=๐›ฝ? Before we consider what this would look like, note that we can rewrite the equation defining this locus. Given that ๐œƒ=๐›ผโˆ’๐›ฝ, we can substitute in the value of ๐›ผ and ๐›ฝ as follows: ๐œƒ=(๐‘งโˆ’๐‘Ž)โˆ’(๐‘งโˆ’๐‘).argarg

Using the properties of the argument, we have ๐œƒ=๏€ป๐‘งโˆ’๐‘Ž๐‘งโˆ’๐‘๏‡.arg

We now consider what shape this locus traces out on the complex plane.

As ๐‘ƒ moves in the plane, the โˆ ๐ด๐‘ƒ๐ต is constant. This property might be familiar from the circle theorems of elementary geometry: recall that the angles subtended from an arc are equal if they are in the same segment. The question is whether the converse of this theorem is also true and, in fact, it is. Therefore, we can conclude that the locus of ๐‘ง which satisfies the given condition traces out the arc of a circle from point ๐ด to point ๐ต. Note that it traces it out in a counterclockwise direction.

Loci Representing Arcs of a Circle

The locus of the point ๐‘ง such that arg๏€ฝ๐‘งโˆ’๐‘ง๐‘งโˆ’๐‘ง๏‰=๐œƒ๏Šง๏Šจ is an arc of a circle which subtends an angle of ๐œƒ between the points represented by ๐‘ง๏Šง and ๐‘ง๏Šจ.

  • If ๐œƒ<๐œ‹2, the locus is a major arc.
  • If ๐œƒ=๐œ‹2, the locus is a semicircle.
  • If ๐œƒ>๐œ‹2, the locus is a minor arc.

Notice that the end points are not part of the locus.

We will now consider an example to demonstrate how we can identify the locus of a point ๐‘ง given the equation that it satisfies.

Example 4: Identifying Loci

Which of following figures is the correct representation of the locus of ๐‘ง which satisfies arg๏€ฝ๐‘งโˆ’5+4๐‘–๐‘ง+2โˆ’4๐‘–๏‰=5๐œ‹18?

Answer

The locus traces out an arc from the points represented by the complex numbers 5โˆ’4๐‘– to โˆ’2+4๐‘– such that it subtends an angle of 5๐œ‹18. Since 5๐œ‹18<๐œ‹2, the locus is a major arc. There are only three choices that represent major arcs: (b), (c), and (d). Secondly, we recall that the locus is swept out in an counterclockwise direction from 5โˆ’4๐‘– to โˆ’2+4๐‘–. This rules out (d) as a possibility. Finally, since the locus does not include the two endpoints, we are looking for the locus that uses two hollow circles, โˆ˜, to represent the end points rather than two solid circles, โ€ข. Hence, the correct representation for the locus of ๐‘ง is figure (b).

Example 5: Finding the Equation of a Locus

The figure shows a locus of a point ๐‘ง in the complex plane. Write an equation for the locus in the form arg๏€ป๐‘งโˆ’๐‘Ž๐‘งโˆ’๐‘๏‡=๐œƒ, where ๐‘Ž,๐‘โˆˆโ„‚ and 0<๐œƒโ‰ค๐œ‹ are constants to be found.

Answer

Recall that if a locus of a point ๐‘ง satisfies arg๏€ป๐‘งโˆ’๐‘Ž๐‘งโˆ’๐‘๏‡=๐œƒ, it is a major arc traced out counterclockwise from the point represented by ๐‘Ž to ๐‘, which subtends an angle of ๐œƒ. Therefore, using the figure, we can immediately see that ๐œƒ=๐œ‹5. We can also easily identify the two endpoints. We need to be careful to ensure that we do not put them in the wrong order. Since the arc it traced out counterclockwise, we can see that the point ๐ด represents the starting point. Hence, ๐‘Ž=4โˆ’3๐‘– and ๐‘=โˆ’3+๐‘–. Therefore, the equation of the locus is given by arg๏€ฝ๐‘งโˆ’(4โˆ’3๐‘–)๐‘งโˆ’(โˆ’3+๐‘–)๏‰=๐œ‹5.

It is possible to use both geometric and algebraic techniques to find the Cartesian equations of loci of this form. In the next example, we will demonstrate geometric methods for finding the equations of loci; then in the following example we will demonstrate algebraic techniques.

Example 6: Finding the Cartesian Equation of a Locus Geometrically

The point ๐‘ง satisfies arg๏€ผ๐‘งโˆ’6๐‘งโˆ’6๐‘–๏ˆ=๐œ‹4.

By plotting the locus on an Argand diagram, find its Cartesian equation.

Answer

The locus is the arc of a circle traced out counterclockwise from the point ๐ด(6,0) to the point ๐ต(0,6) which subtends an angle of ๐œ‹4. Since ๐œ‹4<๐œ‹2, the locus is a major arc.

To find a Cartesian equation, we need to find the equation of the circle; then we restrict the values of ๐‘ฅ and ๐‘ฆ to exclude the minor arc which it not part of the locus. Therefore, we need to find the coordinates of the center. Notice that since the angle at the center of the circle is a right angle, we can see that it lies directly above the point ๐ด(6,0) and on the horizontal that passes through ๐ต(0,6). Hence, its coordinates are (6,6). Additionally, we can immediately see that the radius is equal to 6. Therefore, the Cartesian equation of the circle is given by (๐‘ฆโˆ’6)+(๐‘ฅโˆ’6)=6.๏Šจ๏Šจ๏Šจ

We now need to restrict the values of ๐‘ฅ and ๐‘ฆ so that they are on the locus. Notice that all the values of ๐‘ฅ and ๐‘ฆ lie above the line passing through ๐ด and ๐ต. It is straightforward to see that the equation of this line is ๐‘ฆ=6โˆ’๐‘ฅ. Recall that the locus does not include its end points; therefore, the equation of the locus is given by (๐‘ฆโˆ’6)+(๐‘ฅโˆ’6)=6,๐‘ฆ>6โˆ’๐‘ฅ.๏Šจ๏Šจ๏Šจwhere

Example 7: Finding the Cartesian Equation of a Locus Algebraically

Find the Cartesian equation of the locus of ๐‘ง which satisfies arg๏€ฝ๐‘งโˆ’3๐‘–๐‘งโˆ’5๐‘–๏‰=2๐œ‹3.

Answer

To find the Cartesian equation of a locus, we start by substituting ๐‘ง=๐‘ฅ+๐‘ฆ๐‘– into the complex equation as follows: arg๏€ฝ๐‘ฅ+๐‘ฆ๐‘–โˆ’3๐‘–๐‘ฅ+๐‘ฆ๐‘–โˆ’5๐‘–๏‰=2๐œ‹3.

We will now perform the complex division by multiplying the numerator and denominator by the complex conjugate of the denominator as shown: ๐‘ฅ+๐‘ฆ๐‘–โˆ’3๐‘–๐‘ฅ+๐‘ฆ๐‘–โˆ’5๐‘–=(๐‘ฅ+(๐‘ฆโˆ’3)๐‘–)(๐‘ฅโˆ’(๐‘ฆโˆ’5)๐‘–)(๐‘ฅ+(๐‘ฆโˆ’5)๐‘–)(๐‘ฅโˆ’(๐‘ฆโˆ’5)๐‘–).

Expanding the parentheses in the numerator and denominator gives ๐‘ฅ+๐‘ฆ๐‘–โˆ’3๐‘–๐‘ฅ+๐‘ฆ๐‘–โˆ’5๐‘–=๐‘ฅโˆ’๐‘ฅ๐‘ฆ+5๐‘ฅ๐‘–+๐‘ฅ๐‘ฆโˆ’๐‘ฆโˆ’5๐‘ฆโˆ’3๐‘ฅ๐‘–โˆ’3๐‘ฆ+15๐‘ฅ+(๐‘ฆโˆ’5).๏Šจ๏Šจ๏Šจ๏Šจ

Gathering together the real and imaginary parts, we have ๐‘ฅ+๐‘ฆ๐‘–โˆ’3๐‘–๐‘ฅ+๐‘ฆ๐‘–โˆ’5๐‘–=๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15๐‘ฅ+(๐‘ฆโˆ’5)+2๐‘ฅ๐‘ฅ+(๐‘ฆโˆ’5)๐‘–.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Hence, arg๏€ฟ๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15๐‘ฅ+(๐‘ฆโˆ’5)+2๐‘ฅ๐‘ฅ+(๐‘ฆโˆ’5)๐‘–๏‹=2๐œ‹3.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Using the definition of the argument in terms of the tangent function, we can rewrite this as ๏Šจ๏—๏—๏Šฐ(๏˜๏Šฑ๏Šซ)๏—๏Šฐ๏˜๏Šฑ๏Šฎ๏˜๏Šฐ๏Šง๏Šซ๏—๏Šฐ(๏˜๏Šฑ๏Šซ)๏Žก๏Žก๏Žก๏Žก๏Žก๏Žก=2๐œ‹3.tan

Notice that we need not worry about which quadrant the complex number is in, since adding or subtracting multiples of ๐œ‹ to ๐œƒ has no effect on the value of tan๐œƒ. Simplifying, we have 2๐‘ฅ๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15=โˆ’โˆš3.๏Šจ๏Šจ

Hence, โˆ’โˆš3๏€น๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15๏…=2๐‘ฅ๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15=โˆ’2โˆš33๐‘ฅ๐‘ฅ+2โˆš33๐‘ฅ+๐‘ฆโˆ’8๐‘ฆ+15=0.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

We can now complete the square for both ๐‘ฅ and ๐‘ฆ as follows: ๏€ฟ๐‘ฅ+โˆš33๏‹+(๐‘ฆโˆ’4)โˆ’13โˆ’16+15=0.๏Šจ๏Šจ

Rearranging gives ๏€ฟ๐‘ฅ+โˆš33๏‹+(๐‘ฆโˆ’4)=43.๏Šจ๏Šจ

Finally we need to state the restriction on ๐‘ฅ and ๐‘ฆ to ensure that the points line on the locus, which is the minor arc. Hence, we should only consider the points where ๐‘ฅ>0.

Example 8: Finding the Properties of a Locus

Given that ๐‘ง satisfies arg๏€ผ๐‘งโˆ’3๐‘งโˆ’9๏ˆ=๐œ‹2, by sketching the locus of ๐‘ง, find the range of values of Im(๐‘ง) and the range of values of the principle argument of ๐‘ง.

Answer

Notice that since the value of the argument is ๐œ‹2, we have a semicircle. Furthermore, it starts from the point (3,0) and traces out the arc counterclockwise to the point (9,0). Hence, the line segment between these two points is the diameter and the length of the radius is half this distance; hence ๐‘Ÿ=3.

Since the radius is 3 and the semicircle lies below the ๐‘ฅ-axis, the range of values of Im(๐‘ง) will be from โˆ’3 to 0. In particular, โˆ’3โ‰ค(๐‘ง)<0.Im

Notice that the inequality with zero is strict since the locus does not include its end points. To find the range of values of the argument of ๐‘ง, we need to use some geometric reasoning. Firstly, we draw a ray on the diagram which starts at the origin and is tangent to the semicircle. Using the properties of circles, we know that โˆ ๐ถ๐ต๐ด is a right angle. Furthermore, we know that ๐ต๐ถ=๐‘Ÿ=3 and ๐ด๐ถ=๐‘Ÿ+3=6. Hence, using trigonometry, we can find ๐›ผ=โˆ ๐ต๐ด๐ถ as follows: sin๐›ผ=36.

Hence, ๐›ผ=๐œ‹6.

Therefore, the range of the argument of ๐‘ง is given by โˆ’๐œ‹6โ‰ค(๐‘ง)<0.arg

Notice that the argument is less than zero since the end points are not included in the locus, whereas it includes the argument of the point that is tangent to the circle, so it is greater than or equal to the argument of this point.

Key Points

  1. We can use the argument in the same way that we can use the modulus to define loci in the complex plane.
  2. Using our understanding of geometry and the geometry of the complex plane, we can interpret the loci of points which satisfy certain equations and investigate their properties.
  3. The locus of a point ๐‘ง which satisfies arg(๐‘งโˆ’๐‘ง)=๐œƒ๏Šง is a half-line from, but not including, ๐‘ง๏Šง which makes an angle of ๐œƒ to the horizontal half-line extending from ๐‘ง๏Šง in the positive ๐‘ฅ direction. The angle ๐œƒ is measured in a counterclockwise direction.
  4. The locus of a point ๐‘ง which satisfies arg๏€ฝ๐‘งโˆ’๐‘ง๐‘งโˆ’๐‘ง๏‰=๐œƒ๏Šง๏Šจ is an arc of a circle which subtends an angle of ๐œƒ between the points represented by ๐‘ง๏Šง and ๐‘ง๏Šจ.
    • If ๐œƒ<๐œ‹2, the locus is a major arc.
    • If ๐œƒ=๐œ‹2, the locus is a semicircle.
    • If ๐œƒ>๐œ‹2, the locus is a minor arc.
    The end points are not part of the locus.

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