Lesson Explainer: Complex Number Conjugates Mathematics • 12th Grade

In this explainer, we will learn how to use the properties of conjugate numbers to evaluate an expression.

One important concept related to complex numbers is the idea of the complex conjugate. The idea of a conjugate might not be a new idea. When learning how to manipulate and simplify radicals, you might have been introduced to the idea of a conjugate. For a radical expression π‘Ž+π‘βˆšπ‘, the conjugate is defined as π‘Žβˆ’π‘βˆšπ‘. The complex conjugate is actually a special case of this where 𝑐 is a negative number.

Definition: Complex Conjugate

For a complex number 𝑧=π‘Ž+𝑏𝑖, the complex conjugate, 𝑧, is defined as 𝑧=π‘Žβˆ’π‘π‘–. The complex conjugate is sometimes denoted π‘§βˆ—.

By simple consideration of the definition, we can see that, for any complex number 𝑧, the conjugate of the conjugate is equal to 𝑧; that is, 𝑧=𝑧.

Let us begin with an example where we will find the conjugate of a given complex number.

Example 1: Complex Conjugation

What is the conjugate of the complex number 2βˆ’7𝑖?

Answer

Recall the complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is 𝑧=π‘Žβˆ’π‘π‘–. For the complex number we have been given, π‘Ž=2 and 𝑏=βˆ’7 (we should be careful not to miss the minus sign). Therefore, the complex conjugate is equal to 2βˆ’(βˆ’7)𝑖, which we can simplify to 2+7𝑖.

In the previous example, we found the conjugate of a complex number. Remember that a real number is a special case of a complex number. In the next example, we will consider the conjugate of a real number.

Example 2: The Complex Conjugate of a Real Number

If 𝑧 is a real number, what will its conjugate be equal to?

Answer

The definition of the complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is 𝑧=π‘Žβˆ’π‘π‘–. If 𝑧 is a purely real number, we know that 𝑏=0. This means that if 𝑧 is a real number, 𝑧=𝑧.

Hence, if 𝑧 is a real number, its complex conjugate is the same as the original number.

Similarly, we could ask this question: what is the complex conjugate of a purely imaginary number 𝑧? Using the definition of the complex conjugate, noting that we have π‘Ž=0, we find that, for a purely imaginary number, 𝑧=βˆ’π‘§.

In the next example, we will consider the sum of a complex number with its conjugate.

Example 3: The Sum of a Number and Its Complex Conjugate

Find the complex conjugate of βˆ’7βˆ’π‘– and the sum of this number with its complex conjugate.

Answer

The complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is equal to 𝑧=π‘Žβˆ’π‘π‘–. Hence, for the complex number we have been given, we have π‘Ž=βˆ’7 and 𝑏=βˆ’1. Hence, the complex conjugate is βˆ’7+𝑖.

We can now add these two numbers together and get βˆ’7βˆ’π‘–+βˆ’7+𝑖=βˆ’14.

Notice that the result of adding this number to its complex conjugate was a real number. This is no coincidence: for any complex number 𝑧=π‘Ž+𝑏𝑖, we have 𝑧+𝑧=π‘Ž+𝑏𝑖+π‘Žβˆ’π‘π‘–=2π‘Ž.

We can also write this as 𝑧+𝑧=2(𝑧)Re.

We could also ask this question: what happens when we take the difference of a number with its complex conjugate? In a similar way, we can write π‘§βˆ’π‘§=π‘Ž+π‘π‘–βˆ’(π‘Žβˆ’π‘π‘–)=2𝑏𝑖, which we can also express as π‘§βˆ’π‘§=2𝑖(𝑧)Im.

We will now look at an example in which we consider the difference of a number with its complex conjugate.

Example 4: Solving Equations Involving Complex Conjugates

Find the complex number 𝑧 that satisfies the following equations: 𝑧+𝑧=βˆ’5,π‘§βˆ’π‘§=3𝑖.

Answer

We recall the identity 𝑧+𝑧=2(𝑧)Re. Using this identity, we can write the first equation as 2(𝑧)=βˆ’5.Re

Hence, Re(𝑧)=βˆ’52.

Next, we recall the identity π‘§βˆ’π‘§=2𝑖(𝑧)Im. Before we can apply this identity to the second equation, we first multiply through by βˆ’1, which gives π‘§βˆ’π‘§=βˆ’3𝑖.

Now, we can use the identity to rewrite this as 2𝑖(𝑧)=βˆ’3𝑖.Im

Dividing by 2𝑖 gives Im(𝑧)=βˆ’3𝑖2𝑖=βˆ’32.

This gives us ReandIm(𝑧)=βˆ’52(𝑧)=βˆ’32. Hence, 𝑧=βˆ’52βˆ’32𝑖.

In the next example, we will compute the product of a complex number with its conjugate.

Example 5: The Product of a Complex Number with Its Complex Conjugate

Find the complex conjugate of 1+𝑖 and the product of this number with its complex conjugate.

Answer

For the first part of the question: The complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is equal to 𝑧=π‘Žβˆ’π‘π‘–. We can write the complex number 1+𝑖 as 1+1𝑖, which tells us π‘Ž=1 and 𝑏=1. The complex conjugate of 1+1𝑖 is 1βˆ’1𝑖, which is the same as 1βˆ’π‘–.

Hence, the complex conjugate of 1+𝑖 is 1βˆ’π‘–.

For the second part of the question: Let us consider the product of 1+𝑖 and its conjugate 1βˆ’π‘– by expanding through the parentheses: (1+𝑖)(1βˆ’π‘–)=1+π‘–βˆ’π‘–βˆ’π‘–=1βˆ’π‘–.

Since 𝑖=βˆ’1, the resulting expression is 1βˆ’(βˆ’1)=2.

Hence, the product of 1+𝑖 and its conjugate is equal to 2.

Once again, we notice that the product of this complex number with its complex conjugate is a real number. This is an example of a general property of the complex conjugate. In particular, we find that the product of a complex number and its conjugate is actually a special case of the difference of two squares: (π‘₯βˆ’π‘¦)(π‘₯+𝑦)=π‘₯βˆ’π‘¦οŠ¨οŠ¨. Setting π‘₯=π‘Ž and 𝑦=𝑏𝑖 gives us (π‘Ž+𝑏𝑖)(π‘Žβˆ’π‘π‘–)=π‘Žβˆ’(𝑏𝑖), and using the fact that 𝑖=βˆ’1, we get 𝑧𝑧=π‘Ž+𝑏.

Property: Complex Conjugate

For a complex number 𝑧=π‘Ž+𝑏𝑖,

  • 𝑧+𝑧=2(𝑧)Re,
  • π‘§βˆ’π‘§=2𝑖(𝑧)Im,
  • 𝑧𝑧=π‘Ž+π‘οŠ¨οŠ¨,
  • 𝑧=𝑧 is equivalent to π‘§βˆˆβ„.

In the next example, we will consider how the operation of sum or product of complex numbers interacts with the conjugate operation.

Example 6: Complex Conjugates of Sums and Products

Consider 𝑧=5βˆ’π‘–βˆš3 and 𝑀=√2+π‘–βˆš5.

  1. Calculate 𝑧 and 𝑀.
  2. Find 𝑧+𝑀 and (𝑧+𝑀).
  3. Find 𝑧𝑀 and (𝑧𝑀).

Answer

Part 1

The complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is equal to 𝑧=π‘Žβˆ’π‘π‘–. We can write the complex number 𝑧=5βˆ’π‘–βˆš3 as 5+ο€»βˆ’βˆš3𝑖, which tells us π‘Ž=5 and 𝑏=βˆ’βˆš3, so the complex conjugate of 𝑧 is 𝑧=5βˆ’ο€»βˆ’βˆš3𝑖=5+π‘–βˆš3. Similarly for 𝑀, we can see that π‘Ž=√2 and 𝑏=√5, which leads to the conjugate 𝑀=√2βˆ’π‘–βˆš5.

Part 2

Using the answers from part 1, we calculate 𝑧+𝑀=ο€»5+π‘–βˆš3+ο€»βˆš2βˆ’π‘–βˆš5=5+π‘–βˆš3+√2βˆ’π‘–βˆš5.

By gathering like terms, we can rewrite it as follows: 𝑧+𝑀=5+√2+ο€»βˆš3βˆ’βˆš5𝑖.

Now, we can calculate (𝑧+𝑀). Firstly, we evaluate the term inside the brackets: 𝑧+𝑀=5βˆ’π‘–βˆš3+√2+π‘–βˆš5=5+√2+ο€»βˆš5βˆ’βˆš3𝑖.

To take the complex conjugate of this number, we notice that this number is in the form π‘Ž+𝑏𝑖, where π‘Ž=5+√2 and 𝑏=√5βˆ’βˆš3. Since the complex conjugate takes the form π‘Žβˆ’π‘π‘–, we have (𝑧+𝑀)=5+√2βˆ’ο€»βˆš5βˆ’βˆš3𝑖. Noticing βˆ’ο€»βˆš5βˆ’βˆš3=βˆ’βˆš5+√3=√3βˆ’βˆš5, we can also write this complex number as 5+√2+ο€»βˆš3βˆ’βˆš5𝑖. We can now see that this is the same as the number obtained from the expression 𝑧+𝑀.

Part 3

Using the answers from part 1, we can write 𝑧𝑀=ο€»5+π‘–βˆš3ο‡ο€»βˆš2βˆ’π‘–βˆš5.

Distributing over the parentheses, we get 𝑧𝑀=5√2βˆ’5π‘–βˆš5+π‘–βˆš3√2βˆ’π‘–βˆš3√5.

Gathering like terms and using the fact that 𝑖=βˆ’1, we can rewrite this as 𝑧𝑀=5√2+√15βˆ’ο€»5√5βˆ’βˆš6𝑖.

We calculate (𝑧𝑀) by first finding 𝑧𝑀 and then taking the complex conjugate as follows: 𝑧𝑀=ο€»5βˆ’π‘–βˆš3ο‡ο€»βˆš2+π‘–βˆš5.

Distributing over the parentheses, we find 𝑧𝑀=5√2+5π‘–βˆš5βˆ’π‘–βˆš3√2βˆ’π‘–βˆš3√5.

Simplifying, we get 𝑧𝑀=5√2+√15+ο€»5√5βˆ’βˆš6𝑖.

To take the complex conjugate of this number, we notice that this number is in the form π‘Ž+𝑏𝑖, where π‘Ž=5√2+√15 and 𝑏=5√5βˆ’βˆš6. Since the complex conjugate takes the form π‘Žβˆ’π‘π‘–, we have (𝑧𝑀)=5√2+√15βˆ’ο€»5√5βˆ’βˆš6𝑖.

We can see that this is the same complex number as what we obtained from 𝑧𝑀.

In the previous example, we saw that, for the complex numbers 𝑧 and 𝑀, 𝑧+𝑀=(𝑧+𝑀) and 𝑧𝑀=(𝑧𝑀). This is, in fact, a general rule that holds for any pair of complex numbers, the derivation of which uses exactly the same techniques as used in the previous example.

Property: Algebraic Operations and Complex Conjugates

Given complex numbers 𝑧 and 𝑀, we have the following identities:

  • 𝑧+𝑀=(𝑧+𝑀),
  • π‘§βˆ’π‘€=(π‘§βˆ’π‘€),
  • 𝑧𝑀=(𝑧𝑀),
  • 𝑧𝑀=𝑧𝑀.

In our final example, we will use the properties of the complex conjugate to solve an equation using a complex variable 𝑧.

Example 7: Solving Equations Including Complex Conjugates

Solve 𝑧𝑧+π‘§βˆ’π‘§=4+2𝑖.

Answer

We could approach this problem in one of two ways: we could write 𝑧=π‘Ž+𝑏𝑖 and substitute it into the equation and then solve for π‘Ž and 𝑏; alternatively, we could use the properties of complex conjugates. We will demonstrate both approaches.

Method 1

Recall that the complex conjugate of 𝑧=π‘Ž+𝑏𝑖 is 𝑧=π‘Žβˆ’π‘π‘–. Using this expression, we have 4+2𝑖=𝑧𝑧+π‘§βˆ’π‘§=(π‘Ž+𝑏𝑖)(π‘Žβˆ’π‘π‘–)+(π‘Žβˆ’π‘π‘–)βˆ’(π‘Ž+𝑏𝑖).

We can first distribute (π‘Ž+𝑏𝑖)(π‘Žβˆ’π‘π‘–)=π‘Žβˆ’π‘Žπ‘π‘–+π‘Žπ‘π‘–βˆ’π‘π‘–, which simplifies to π‘Ž+π‘οŠ¨οŠ¨ when remembering that 𝑖=βˆ’1.

Distributing over the remaining parentheses, while being careful to notice βˆ’(π‘Ž+𝑏𝑖)=βˆ’π‘Žβˆ’π‘π‘–, we have 4+2𝑖=π‘Ž+𝑏+π‘Žβˆ’π‘π‘–βˆ’π‘Žβˆ’π‘π‘–=π‘Ž+π‘βˆ’2𝑏𝑖.

This leads to the equation 4+2𝑖=π‘Ž+π‘βˆ’2𝑏𝑖. Recall that the two complex numbers are equal when the real and the imaginary parts are both equal.

Equating the real parts of the complex numbers on both sides of the equation gives 4=π‘Ž+𝑏. Equating the imaginary parts gives 2=βˆ’2𝑏, which leads to 𝑏=βˆ’1. Substituting in the value for 𝑏 in the equation above, we get π‘Ž+(βˆ’1)=4π‘Ž=3.

Hence, π‘Ž=±√3. This gives us that π‘Ž could be either √3 or βˆ’βˆš3, while 𝑏 is equal to βˆ’1. Since 𝑧=π‘Ž+𝑏𝑖, we have two possible solutions to the equation: 𝑧=√3βˆ’π‘–π‘§=βˆ’βˆš3βˆ’π‘–.and

Method 2

Let us use the properties of complex conjugates as follows. We first notice that the left-hand side of the equation has two parts:

For each of these two parts, we recall the identities

  • 𝑧𝑧=π‘Ž+π‘οŠ¨οŠ¨,
  • π‘§βˆ’π‘§=2𝑖(𝑧)Im.

To obtain the expresion (2), we can multiply both sides of the second identity to obtain π‘§βˆ’π‘§=βˆ’2𝑖(𝑧).Im Since Im(𝑧)=𝑏, we can obtain (2)=βˆ’2𝑏𝑖. For (1), the first identity gives (1)=π‘Ž+π‘οŠ¨οŠ¨. Substituting these expressions into the equation above, we obtain π‘Ž+π‘βˆ’2𝑏𝑖=4+2𝑖.

This leads us to the same equation as what we derived using the first method.

Using the same computations, the solutions to the given equations are 𝑧=√3+𝑖,βˆ’βˆš3+𝑖.and

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • For a complex number 𝑧=π‘Ž+𝑏𝑖, its complex conjugate is defined as 𝑧=π‘Žβˆ’π‘π‘–.
  • For two complex numbers 𝑧=π‘Ž+π‘π‘–οŠ§ and 𝑧=𝑐+π‘‘π‘–οŠ¨, the following identities hold:
    • (𝑧±𝑧)=π‘§Β±π‘§οŠ§οŠ¨οŠ§οŠ¨,
    • (𝑧𝑧)=π‘§π‘§οŠ§οŠ¨οŠ§οŠ¨,
    • 𝑧+𝑧=2(𝑧)Re,
    • π‘§βˆ’π‘§=2𝑖(𝑧)Im,
    • 𝑧𝑧=π‘Ž+π‘οŠ§οŠ§οŠ¨οŠ¨.
  • A complex number is equal to its conjugate if and only if it is a real number.

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