Explainer: Argand Diagram

In this explainer, we will learn how to identify complex numbers plotted on an Argand diagram and discover their geometric properties.

One of the most fascinating things about complex numbers is that they introduce a geometric interpretation to familiar arithmetic operations. When working with purely real numbers, we could express them on a one-dimensional number line. Thinking in this way gave us additional insight into their properties. Whereas, with the introduction of ๐‘–, we can add a second dimension and consider complex numbers as points in a plane, we will find that visualizing complex numbers in this way will give us additional insight into their properties.

Definition: Argand Diagram

The complex numbers can be represented geometrically on a two-dimensional plane with two perpendicular axes representing the real and imaginary parts of the number respectively. The complex number ๐‘ง=๐‘ฅ+๐‘ฆ๐‘– is represented by the point (๐‘ฅ,๐‘ฆ) in Cartesian coordinates. This plane is referred to as the complex plane, the Argand plane, or the Argand diagram.

Example 1: Coordinates of Complex Numbers on an Argand Diagram

If the number ๐‘=8+๐‘– is represented on an Argand diagram by the point ๐ด, determine the Cartesian coordinates of that point.

Answer

From the definition of the Argand diagram, we know that the complex number ๐‘ง=๐‘Ž+๐‘๐‘– will be represented by a point with Cartesian coordinates (๐‘Ž,๐‘). Hence, ๐‘ will be represented by the point ๐ด(8,1).

Example 2: Representing Complex Numbers on an Argand Diagram

Seven complex numbers ๐‘ง1, ๐‘ง2, ๐‘ง3, ๐‘ง4, ๐‘ง5, ๐‘ง6, and ๐‘ง7 are represented on the Argand diagram.

  1. Which of the complex numbers is โˆ’3+2๐‘–?
  2. What complex number is represented by ๐‘ง4?
  3. Which complex number has equal real and imaginary parts?
  4. Which two complex numbers are a conjugate pair? What is their geometric relationship?

Answer

Part 1

According to the definition of the Argand diagram, the complex number โˆ’3+2๐‘– will be represented by the point (โˆ’3,2). Reading these coordinates off the plane, we find that โˆ’3+2๐‘–=๐‘ง3.

Part 2

We begin by reading off the coordinates of ๐‘ง4 from the Argand diagram as (โˆ’4,โˆ’1) which according to the definition represent the complex number โˆ’4โˆ’๐‘–. Hence, ๐‘ง3=โˆ’4โˆ’๐‘–.

Part 3

A complex number with equal real and imaginary parts will lie on the line ๐‘ฅ=๐‘ฆ. Drawing this line on the Argand diagram, we find that only one of the numbers lies on this line: ๐‘ง5.

Part 4

Recall that the complex conjugate of ๐‘ง=๐‘Ž+๐‘๐‘– is ๐‘งโˆ—=๐‘Žโˆ’๐‘๐‘–. Therefore, we could plot ๐‘ง at the point (๐‘Ž,๐‘) and we could plot ๐‘งโˆ— at the point (๐‘Ž,โˆ’๐‘). Hence, the points representing a complex number and its conjugate both have the same ๐‘ฅ-value but opposite ๐‘ฆ-values. Looking at the diagram we have been given, we see there are only two pairs of points with the same ๐‘ฅ-coordinates: ๐‘ง2 and ๐‘ง5 and ๐‘ง1 and ๐‘ง6. Considering ๐‘ง2 and ๐‘ง5, we find that the ๐‘ฆ-coordinate of ๐‘ง2 is 3 whereas the ๐‘ฆ-coordinate of ๐‘ง5 is โˆ’2. Hence, these two are not a complex conjugate pair, whereas, considering ๐‘ง1 and ๐‘ง6, we find that the ๐‘ฆ-coordinate of ๐‘ง1 is 3 and the ๐‘ฆ-coordinate of ๐‘ง6 is โˆ’3. Therefore, they are a complex conjugate pair. Furthermore, we can see that, as a complex conjugate pair, the points ๐‘ง1 and ๐‘ง6 are related by reflection in the real axis (๐‘ฅ-axis).

Using Argand diagrams, we can interpret addition of complex numbers geometrically. For two complex numbers ๐‘ง1=๐‘Ž+๐‘๐‘– and ๐‘ง2=๐‘+๐‘‘๐‘–, their sum can be expressed as ๐‘ง1+๐‘ง2=(๐‘Ž+๐‘)+(๐‘+๐‘‘)๐‘–. If we plot these numbers on the Argand diagram, we would plot the points (๐‘Ž,๐‘), (๐‘,๐‘‘), and (๐‘Ž+๐‘,๐‘+๐‘‘). Considering these points suggests some sort of equivalence between complex numbers and vectors. This is in fact true, and for a number of operations with complex numbers, considering them to be vectors in the Argand diagram is actually most informative. In particular, for addition and subtraction, we can consider the two complex numbers ๐‘ง1 and ๐‘ง2 to represented vectors with components โŸจ๐‘Ž,๐‘โŸฉ, โŸจ๐‘,๐‘‘โŸฉ respectively. In this way, addition of complex numbers can be interpreted as vector addition. For example, adding the complex numbers 1+2๐‘– and 3+๐‘– using the parallelogram rule can be represented as follows.

We can also use geometric interpretations to understand mappings on the complex plane as the next example will demonstrate.

Example 3: Geometric Interpretation of Addition

Describe the geometric transformation that occurs when numbers in the complex plane are mapped to their sum with 3โˆ’2๐‘–.

Answer

To answer this question, it is helpful to interpret complex numbers as vectors in the Argand diagram. In this way, addition of complex numbers can be interpreted as vectors addition whereby, adding one vector, A, to another, B, can be understood to be a translation of vector A to the tip of vector B or vice versa. Therefore, a transformation that takes ๐‘ง to ๐‘ง+3โˆ’2๐‘– can be understood to represents a translation of the complex plane by the vector โŸจ3,โˆ’2โŸฉ.

Interpreting complex numbers as vectors can also be helpful to give a geometric interpretation to multiplying a complex number by a real number as the next example will demonstrate.

Example 4: Geometric Interpretation of Multiplication by a Real Number

Three complex numbers ๐‘ง1, ๐‘ง2, and ๐‘ง3 are represented on the Argand diagram.

  1. Find the image of points ๐‘ง1, ๐‘ง2, and ๐‘ง3 under the transformation that maps ๐‘ง to 2๐‘ง.
  2. By plotting these points on an Argand diagram, or otherwise, give a geometric interpretation of the transformation.

Answer

Part 1

We begin by finding the values ๐‘ง1, ๐‘ง2, and ๐‘ง3 from their points on the Argand diagram. Since ๐‘ง1 is at point (2,0), its value must be ๐‘ง1=2. Similarly, ๐‘ง2 is at (0,โˆ’1); hence, its value will be ๐‘ง2=โˆ’๐‘–, and ๐‘ง3 is at (โˆ’3,2); therefore, its value will be ๐‘ง3=โˆ’3+2๐‘–. To find their images under the map that takes ๐‘ง to 2๐‘ง, we simply need to multiply by 2. Hence, ๐‘งโ€ฒ1=2ร—2=4,๐‘งโ€ฒ2=2ร—โˆ’๐‘–=โˆ’2๐‘–,๐‘งโ€ฒ3=2ร—(โˆ’3+2๐‘–)=โˆ’6+4๐‘–.

Part 2

The plot of these points on an Argand diagram is show below.

Visually, we can see that each point is mapped to a point that is further from the origin. In fact, thinking about complex numbers as vectors, we can see that each vector is mapped to another vector in the same direction with twice the length. Therefore, the map represents a dilation with scale factor two centered at the origin.

Interpreting complex numbers as vectors in the Argand diagram enabled us to interpret multiplication by a real number ๐‘ as a dilation with scale factor ๐‘ centered at the origin. This interpretation also applies to negative numbers, where you have a dilation by a negative scale factor ๐‘. Negative scale factor dilations can also be interpreted as a rotation by ๐œ‹ radians followed by a dilation with scale factor |๐‘|.

We will now turn our attention to the geometric interpretation of multiplying by an imaginary number.

Example 5: Geometric Interpretation of Multiplication by ๐‘–

Four complex numbers ๐‘ง1, ๐‘ง2, ๐‘ง3, and ๐‘ง4 are shown on the Argand diagram.

  1. Find the image of the points ๐‘ง1, ๐‘ง2, ๐‘ง3, and ๐‘ง4 under a transformation that maps ๐‘ง to ๐‘–๐‘ง.
  2. By plotting these points on an Argand diagram, or otherwise, give a geometric interpretation of the transformation.

Answer

Part 1

We begin by finding the values of ๐‘ง1, ๐‘ง2, ๐‘ง3, and ๐‘ง4 form their points on the Argand diagram as shown in the table below.

PointValue
๐‘ง1(3,0)๐‘ง1=3
๐‘ง2(2,3)๐‘ง2=2+3๐‘–
๐‘ง3(โˆ’2,โˆ’1)๐‘ง3=โˆ’2โˆ’๐‘–
๐‘ง4(0,โˆ’1)๐‘ง4=โˆ’๐‘–

Now we multiply each complex number by ๐‘– to find the image under the transformation as shown follows: ๐‘งโ€ฒ1=๐‘–ร—3=3๐‘–,๐‘งโ€ฒ2=๐‘–ร—(2+3๐‘–)=2๐‘–+3๐‘–2=โˆ’3+2๐‘–,๐‘งโ€ฒ3=๐‘–ร—(โˆ’2โˆ’๐‘–)=โˆ’2๐‘–โˆ’๐‘–2=1โˆ’2๐‘–,๐‘งโ€ฒ4=๐‘–ร—โˆ’๐‘–=โˆ’๐‘–2=1.

Part 2

Considering these complex numbers as vectors on the Argand diagram, we plot each complex number with its vector. We have also included an arrow indicating the transformation under this map.

We can see that the image of each point is the same distance from the origin but at a different angle with the real axis. When we consider all points together, we see that they are all rotated by the same angle. Specifically, the transformation represents a rotation by an angle of ๐œ‹2 radians counterclockwise about the origin.

The previous example demonstrates that multiplying by ๐‘– represents a counterclockwise rotation by ๐œ‹2 radians about the origin. Combining this with the geometric interpretation of multiplying by a real number, we can interpret multiplying by a general imaginary number as a rotation followed by a dilation. Considering โˆ’1 to be the product ๐‘–ร—๐‘–, we can think of the transformation representing multiplicaion by โˆ’1 as the transformation we get by multiplying by ๐‘– twice. So mulitplication by โˆ’1 is the transformation we get by rotating counterclockwise by ๐œ‹2 twice; that is, it is a rotation by ๐œ‹ radians. This is the same as the interpretation that multiplying by โˆ’1 represents a dilation with a scale factor of โˆ’1.

To gain a geometric understanding of multiplication and division by arbitrary complex numbers, we will need first to gain an understanding of the modulus and argument of complex numbers.

We finish this explainer by looking at one final example which demonstrates some of the interesting geometric insights that can be gained by considering complex numbers in the Argand diagram.

Example 6: Geometric Interpretation of the Roots of Unity

  1. Find all the solutions to ๐‘ง6=1.
  2. By plotting the solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions of ๐‘ง6=1.

Answer

Part 1

Given that ๐‘ง is raised to the sixth power, we would expect there to be six roots to the equation. By simple inspection, we can see that both 1 and โˆ’1 are roots. However, to find all the roots, we need to have a look to see whether we can factor the equation into a form we can work with. We start by rewriting the equation in the form ๐‘ง6โˆ’1=0.

At this point, we can see that this is actually a special case of the difference of two squares ๐‘Ž2โˆ’๐‘2=(๐‘Žโˆ’๐‘)(๐‘Ž+๐‘), where ๐‘Ž=๐‘ง3 and ๐‘=1. Therefore, we can rewrite the equation in the form

๏€น๐‘ง3โˆ’1๏…๏€น๐‘ง3+1๏…=0.(1)

We can now consider each factor separately. Starting with the first, we can see that ๐‘ง=1 is one solution. Hence, we will be able to express ๐‘ง3โˆ’1=(๐‘งโˆ’1)๏€น๐‘Ž๐‘ง2+๐‘๐‘ง+๐‘๏….

By expanding the right-hand side, we have ๐‘ง3โˆ’1=๐‘Ž๐‘ง3+๐‘๐‘ง2+๐‘๐‘งโˆ’๐‘Ž๐‘ง2โˆ’๐‘๐‘งโˆ’๐‘. Gathering like terms on the right-hand side, we have ๐‘ง3โˆ’1=๐‘Ž๐‘ง3+(๐‘โˆ’๐‘Ž)๐‘ง2+(๐‘โˆ’๐‘)๐‘งโˆ’๐‘. By equating coefficients, we can immediately see that ๐‘Ž=1 and ๐‘=1. Additionally, ๐‘โˆ’๐‘Ž=0 and ๐‘โˆ’๐‘=0; substituting in the values of ๐‘Ž and ๐‘, we find that ๐‘=1. Hence, we have 0=๐‘ง3โˆ’1=(๐‘งโˆ’1)๏€น๐‘ง2+๐‘ง+1๏…. We can now find the roots of the quadratic equation ๐‘ง2+๐‘ง+1. We will solve this using the quadratic formula ๐‘ง=โˆ’๐‘ยฑโˆš๐‘2โˆ’4๐‘Ž๐‘2๐‘Ž. Substituting in the value ๐‘Ž=๐‘=๐‘=1, we have ๐‘ง=โˆ’1ยฑโˆš12โˆ’42, simplifying, we have ๐‘ง=โˆ’12ยฑโˆš32๐‘–. Similarly, we can consider the second factor ๏€น๐‘ง3+1๏… from equation (1). Once again, by inspection, we can see that ๐‘ง=โˆ’1 is one of the solutions. Therefore, we will be able to rewrite ๐‘ง3+1=(๐‘ง+1)๏€น๐‘Ž๐‘ง2+๐‘๐‘ง+๐‘๏…. Using the same method as above, we find ๐‘Ž=1, ๐‘=โˆ’1, and ๐‘=1. Hence, we have 0=๐‘ง3+1=(๐‘ง+1)๏€น๐‘ง2โˆ’๐‘ง+1๏…. We can now find the roots of the quadratic equation ๐‘ง2โˆ’๐‘ง+1. Using the quadratic formula with ๐‘Ž=๐‘=1 and ๐‘=โˆ’1, we have ๐‘ง=โˆ’(โˆ’1)ยฑโˆš(โˆ’1)2โˆ’42, simplifying, we get ๐‘ง=12ยฑโˆš32๐‘–. In summary, we have found six solutions: ๐‘ง=1,โˆ’1,12+โˆš32๐‘–,12โˆ’โˆš32๐‘–,โˆ’12+โˆš32๐‘–,โˆ’12โˆ’โˆš32๐‘–.Part 2

Plotting these points on an Argand diagram, we find that all the solutions are evenly spaced around the unit circle centered at the origin.