Lesson Explainer: Transformations of the Complex Plane Mathematics

In this explainer, we will learn how to translate and rotate a complex number in the complex plane.

When we consider transformation of the complex plane, we are thinking of functions that map a two-dimensional space to a two-dimensional space. On first impression, it can be difficult to understand how to think about and interpret such functions. The difficulty arises from the fact that the most common techniques we use to understand real functions tend to fall down when we try to understand functions mapping two dimensions to two dimensions. This explainer will introduce a number of techniques we use to understand complex functions.

We will begin by looking at some simple examples to introduce the techniques we use.

Example 1: Translations of the Complex Plane

Find an equation for the image of |𝑧|=2 under the transformation of the complex plane π‘‡βˆΆπ‘§β†¦π‘§+1+𝑖.

Answer

We are considering a transformation 𝑇 from the 𝑧-plane to another complex plane which we will call the 𝑀-plane. We can then represent the transformation as 𝑀=𝑧+1+𝑖.

We can find an equation for the image of |𝑧|=2 by first expressing 𝑧 in terms of 𝑀 as follows: 𝑧=π‘€βˆ’1βˆ’π‘–.

At this point, we can substitute this into the equation |𝑧|=2 which yields |π‘€βˆ’1βˆ’π‘–|=2.

This represents a circle of radius 2 centered at the point represented by the complex number 1+𝑖. We can represent both the 𝑧-plane and the 𝑀-plane as follows.

The previous example introduces us to one of the key techniques we use to understand transformations of the complex plane: considering the image of different curves under the transformation. Understanding the effect of a given transformation on circles and lines will often give us a reasonable impression of what the transformation is doing. More generally, looking at the effect of a transformation on curves and regions is a technique we use to help visualize more complex types of transformation.

Additionally, the previous example demonstrated that a transformation in the form π‘‡βˆΆπ‘§β†¦π‘§+π‘§οŠ§, where 𝑧=π‘₯+π‘¦π‘–οŠ§ is constant (sometimes described as the transformation 𝑀=𝑧+π‘§οŠ§), represents a translation by the vector ο€»π‘₯𝑦.

In the next example, we will consider transformations in the form 𝑀=π‘˜π‘§ where π‘˜βˆˆβ„.

Example 2: Dilations of the Complex Plane

Find an equation for the image of |𝑧|=1 under the transformation of the complex plane π‘‡βˆΆπ‘§β†¦12𝑧.

Answer

We are considering a transformation 𝑇 from the 𝑧-plane to another complex plane which we will call the 𝑀-plane. We can represent this transformation as 𝑀=12𝑧.

We can find an equation for the image of |𝑧|=1 by first expressing 𝑧 in terms of 𝑀 as follows: 𝑧=2𝑀.

At this point, we can substitute this into the equation |𝑧|=1 which yields |2𝑀|=1.

Using the properties of the modulus, we can rewrite this as |2||𝑀|=1, which is equivalent to |𝑀|=12.

This represents a circle of radius 12 centered at the origin. We can represent both the 𝑧-plane and the 𝑀-plane as follows.

The previous example demonstrated that transformations of the form π‘‡βˆΆπ‘§β†¦π‘˜π‘§ where π‘˜βˆˆβ„ is constant are dilations by a scale factor of π‘˜. Although this is a true statement, strictly speaking, the previous example is not sufficient to demonstrate this. We saw that these transformations map circles centered at the origin to other circles centered at the origin with a scaled radius. However, it could be possible that the transformation has also rotated points in the plane. Although in this case this is not true, it does demonstrate that generally we should consider the images of both points and circles to ensure we have a better picture of what a transformation is doing.

Example 3: Rotations of the Complex Plane

Find an equation for the image of arg(𝑧)=πœƒ under the transformation π‘‡βˆΆπ‘§β†¦π‘’π‘§οƒοŽŠ that maps the 𝑧-plane to 𝑀-plane.

Answer

We can find an equation for the image of arg(𝑧)=πœƒ by first expressing 𝑧 in terms of 𝑀 as follows: 𝑧=𝑀𝑒.οŠ±οƒοŽŠ

We can now substitute this into the equation arg(𝑧)=πœƒ which yields arg𝑀𝑒=πœƒ.οŠ±οƒοŽŠ

Using the properties of the argument, we can rewrite this as argarg(𝑀)+𝑒=πœƒ.οŠ±οƒοŽŠ

Since arg𝑒=βˆ’πœ‘οŠ±οƒοŽŠ, we can simplify this to arg(𝑀)=πœƒ+πœ‘.

This represents a half-line whose end point is at the origin, which makes an angle of πœƒ+πœ‘ with the positive real axis.

The previous example showed us that, under the transformation 𝑀=π‘’π‘§οƒοŽŠ, the image half-line at the origin which makes an angle of πœƒ to the real axis is a half-line at the origin which makes an angle of πœƒ+πœ‘. This demonstrates that the transformation rotated the plane by πœ‘. Moreover, we know that it did not scale the line since the modulus of π‘’οƒοŽŠ is one.

Notice that had we considered the image of a circle in the previous example, we would have found that its image would be a circle in the 𝑀-plane with the same center and radius. This would not have been very insightful. This is why we often need to consider the image of both lines and circles to gain a clearer understanding of what a given transformation is doing.

Combining the results of examples 2 and 3, we can see that multiplication by a general complex number 𝑧=π‘Ÿπ‘’οŠ§οƒοΌ can be understood as a combination of the transformation we get by multiplying by the real number π‘Ÿ, which is a dilation by scale factor π‘Ÿ, and the transformation we get by multiplying by 𝑒, which is a counterclockwise rotation about the origin through an angle of πœƒ.

Basic Transformation of the Complex Plane

Let 𝑧=π‘₯+π‘¦π‘–οŠ§ where π‘₯,π‘¦βˆˆβ„ are constant.

  1. The transformation π‘‡βˆΆπ‘§β†¦π‘§+π‘§οŠ§οŠ§ represents a translation by the vector ο€»π‘₯𝑦.
  2. The transformation π‘‡βˆΆπ‘§β†¦π‘§π‘§οŠ¨οŠ§ represents a dilation by scale factor |𝑧| and a counterclockwise rotation about the origin by arg(𝑧).

More involved transformations can be built up by combining these transformations. For example, we can understand the transformation of the form π‘‡βˆΆπ‘§β†¦π‘§π‘§+π‘§οŠ¨οŠ§ as a composition of the transformations π‘‡βˆΆπ‘§β†¦π‘§π‘§οŠ§οŠ¨ and π‘‡βˆΆπ‘§β†¦π‘§+π‘§οŠ¨οŠ§. That is, we can understand it to be a dilation by scale factor |𝑧| with a counterclockwise rotation about the origin by arg(𝑧), followed by a translation by the vector representing the complex number π‘§οŠ§.

In the next couple of examples, we will consider composite transformation of this form.

Example 4: Compound Transformations of the Complex Plane

Find an equation for the image of the half-line arg(π‘§βˆ’4+5𝑖)=πœ‹2 under the transformation π‘‡βˆΆπ‘§β†¦π‘–π‘§βˆ’3βˆ’π‘–.

Answer

We are considering a transformation 𝑇 from the 𝑧-plane to another complex plane which we will call the 𝑀-plane. We can represent this transformation as 𝑀=π‘–π‘§βˆ’3βˆ’π‘–.

We can find an equation for the image of arg(π‘§βˆ’4+5𝑖)=πœ‹2 by first expressing 𝑧 in terms of 𝑀 as follows: 𝑧=𝑀+3+𝑖𝑖.

Substituting this into the equation arg(π‘§βˆ’4+5𝑖)=πœ‹2 yields arg𝑀+3+π‘–π‘–βˆ’4+5𝑖=πœ‹2.

Expressing the subject of the argument as a single fraction, we have arg𝑀+3+π‘–βˆ’4𝑖+5π‘–π‘–οŠ=πœ‹2 which we can simplify to argο€½π‘€βˆ’2βˆ’3𝑖𝑖=πœ‹2.

Using the properties of the argument, we can rewrite this as argarg(π‘€βˆ’2βˆ’3𝑖)βˆ’(𝑖)=πœ‹2.

Since arg(𝑖)=πœ‹2, we have arg(π‘€βˆ’2βˆ’3𝑖)=πœ‹.

This represents a half-line whose end point is at 3+3𝑖 which makes an angle of πœ‹ with the positive horizontal. We can interpret this transformation as the combination of two transformations π‘‡οŠ§, a counterclockwise rotation about the origin by πœ‹2 radians, and π‘‡οŠ¨, a translation by the vector ο€Όβˆ’3βˆ’1. We can represent this visually as follows.

Example 5: Composite Transformations of the Complex Plane

Find an equation for the image of |π‘§βˆ’2|=3 under the transformation of 𝑧-plane to the 𝑀-plane given by 𝑀=𝑖(2𝑧+2).

Answer

We begin by expressing 𝑧 in terms of 𝑀 as follows: 𝑧=βˆ’π‘–π‘€βˆ’22.

We can now substitute this into the equation |π‘§βˆ’2|=3 to find the equation (in terms of 𝑀) for the image of this locus under the given transformation. Hence, |||βˆ’π‘–π‘€βˆ’22βˆ’2|||=3.

Expressing the subject of the modulus as a single fraction, we have |||βˆ’π‘–π‘€βˆ’62|||=3.

Factoring out βˆ’π‘– from the numerator, we have |||βˆ’π‘–2(π‘€βˆ’6𝑖)|||=3.

We can now use the properties of the modulus to rewrite this as |||βˆ’π‘–2||||π‘€βˆ’6𝑖|=3.

Since |||βˆ’π‘–2|||=12, we can express this as |π‘€βˆ’6𝑖|=6.

This is a circle of radius 6 centered at 6𝑖. We can interpret this transformation as the combination of three transformations: π‘‡οŠ§, a dilation centered at the origin with scale factor 2, followed by π‘‡οŠ¨, a translation by the vector ο€Ό20, followed by π‘‡οŠ©, a counterclockwise rotation about the origin by πœ‹2 radians. We can represent this visually as follows.

In the final example, we will consider a transformation which, unlike the transformations we have considered thus far, does not map straight lines to straight lines, and consequently, it is more challenging to gain an understanding of the effect of the transformation. However, using the techniques we have learned we can consider its effect on lines and circles and, as a result, start to build up a picture of what the transformation is doing.

Example 6: Transformations of the Complex Plane

A transformation which maps the 𝑧-plane to the 𝑀-plane is given by 𝑀=π‘§οŠ¨.

  1. Find a Cartesian equation for the image of |𝑧|=8 under the transformation.
  2. Find a Cartesian equation for the image of Re(𝑧)=4.
  3. Find a Cartesian equation for the image of Im(𝑧)=4.

Answer

For the points in the 𝑧-plane, we will denote the real and imaginary parts as π‘₯ and 𝑦 respectively, whereas, for the 𝑀 plane, we will denote them 𝑒 and 𝑣.

Part 1

We will first find a complex equation for the image of |𝑧|=8 in the 𝑀-plane; then we will transform it to a Cartesian equation in terms of 𝑒 and 𝑣. We define 𝑀=π‘§οŠ¨. Taking the modulus of both sides of this equation, we have |𝑀|=|𝑧|.

Using the properties of the modulus, we can rewrite this as |𝑀|=|𝑧|=|𝑧||𝑧|.

Given the fact that we would like to find the image of |𝑧|=8, we can simply replace |𝑧| with 2 in the equation above which gives |𝑀|=64.

Hence, the image of |𝑧|=8 under the transformation is |𝑀|=64. Since this is a circle of radius 64 centered at the origin, we can give the Cartesian equation of this as 𝑒+𝑣=64.

We can visualize this as follows.

Part 2

The points in the 𝑧-plane which satisfy Re(𝑧)=4 can all be written in the form 𝑧=4+𝑦𝑖. To consider the image of this under the transformation 𝑀=π‘§οŠ¨, we substitute this into the equation as follows: 𝑀=(4+𝑦𝑖).

Expanding the parentheses, we have 𝑀=(16βˆ’π‘¦)+8𝑦𝑖.

Since 𝑀=𝑒+𝑣𝑖, we can equate the real and imaginary parts to get the two simultaneous equations 𝑒=16βˆ’π‘¦,𝑣=8𝑦.

To get a Cartesian equation from these two equations, we would like to eliminate 𝑦. From the second equation, we have 𝑦=𝑣8. We can substitute this into the first equation to get 𝑒=16βˆ’ο€»π‘£8=16βˆ’π‘£64.

We can visualize this as follows.

Part 3

All of the points in the 𝑧-plane which satisfy Im(𝑧)=4 can be written in the form 𝑧=π‘₯+4𝑖. Substituting this into the equation 𝑀=π‘§οŠ¨, we have 𝑀=(π‘₯+4𝑖).

Expanding the parentheses, we have 𝑀=ο€Ήπ‘₯βˆ’16+8π‘₯𝑖.

Since 𝑀=𝑒+𝑣𝑖, we can equate the real and imaginary parts to get the two simultaneous equations 𝑒=π‘₯βˆ’16,𝑣=8π‘₯.

To get a Cartesian equation form these two equations, we would like to eliminate π‘₯. From the second equation, we have π‘₯=𝑣8. We can substitute this into the first equation to get 𝑒=𝑣8ο‡βˆ’16=𝑣64βˆ’16.

We can visualize this as follows.

As we have seen in the previous example, understanding a transformation’s effect on lines and circles helps us gain an understanding of its overall effect on the complex plane, even in the case of more general transformations. There are many other techniques which are useful for gaining this understanding, for example, considering the preimage of lines and circles, color maps, vector fields, and discovering the fixed points of the transformation.

Key Points

  • We can gain an understanding of transformations in the complex plane by considering their effect on lines and circles.
  • For a complex number 𝑧=π‘₯+π‘¦π‘–οŠ§, where π‘₯,π‘¦βˆˆβ„, the transformation π‘‡βˆΆπ‘§β†¦π‘§+π‘§οŠ§οŠ§ represents a translation by the vector ο€»π‘₯𝑦.
  • The transformation π‘‡βˆΆπ‘§β†¦π‘§π‘§οŠ¨οŠ§ represents a dilation by scale factor |𝑧| and a counterclockwise rotation about the origin by arg(𝑧).
  • A composite transformation can be built up by composing multiple transformations. We can understand their affect by considering the effect of each transformation in turn.
  • To gain insight into more complex transformations such as 𝑇(𝑧)=π‘§οŠ¨, we can also consider how it maps lines and circles. However, understanding the exact nature of the transformation will generally be more challenging.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.