Explainer: Pure Imaginary Numbers

In this explainer, we will learn how to evaluate, simplify, and multiply pure imaginary numbers and solve equations over the set of pure imaginary numbers.

Gaining confidence in working with imaginary numbers will enable us to require the necessary skills to work effectively with complex numbers more generally.

Historically, the introduction of complex numbers was primarily associated with the idea of solving equations. In particular, in the 16th century, mathematicians were working to find algebraic solutions to the cubic equation. Intriguingly, the equations mathematicians were trying to solve often had purely real solutions. However, the methods required to solve them ended up with the need to evaluate the square roots of negative numbers. In particular, Tartaglia’s method for solving cubics of the form π‘₯+𝑝π‘₯+π‘ž=0 often led to the need to evaluate the square root of negative numbers even when the solutions were all real. For example, applying his method to the equation π‘₯βˆ’π‘₯=0 results in the following solution: 1√3ο€Ώο€»βˆšβˆ’1+ο€»βˆšβˆ’1.οŽͺ

However, by inspection, we can see the equation π‘₯βˆ’π‘₯=0 has three real solutions: 0, 1, and βˆ’1. At the time, many people would have dismissed an expression like this as nonsensical. However, the mathematician Rafael Bombelli saw the usefulness of working with the square roots of negative numbers and, as a result, today we credit him as the first person to formalize their properties.

Let us recap the definition of imaginary numbers.

Definition: Imaginary Numbers

The number 𝑖 is defined as the solution to the equation π‘₯=βˆ’1. Since 𝑖 is not a real number, it is referred to as an imaginary number and all real multiples of 𝑖 (numbers of the form 𝑏𝑖, where 𝑏 is real) are called (purely) imaginary numbers. Often 𝑖 is referred to as the square root of negative one.

As mentioned, the introduction of imaginary numbers enables us to solve equations which have no real solutions. We begin by looking at a simple example of this.

Example 1: Solving Equations Using Imaginary Numbers

Solve the equation 2π‘₯=βˆ’50.

Answer

We begin with isolating π‘₯ by dividing both sides of the equation by two: π‘₯=βˆ’25.

Taking the square root of both sides, we get π‘₯=Β±βˆšβˆ’25, remembering that in taking the square root we need to consider both the positive and the negative solutions. We can rewrite βˆšβˆ’25=√25Γ—βˆ’1=√25Γ—βˆšβˆ’1.

Hence, π‘₯=Β±5𝑖.

Substituting this back into the equation, we can check our answer. Here we check our answer for βˆ’5𝑖: 2π‘₯=2(βˆ’5𝑖)=2(βˆ’5)𝑖.

Since 𝑖=βˆ’1, we can rewrite this as 2π‘₯=2Γ—25Γ—(βˆ’1)=βˆ’50 as required.

By applying the familiar rules of arithmetic and algebra, we can learn to easily work with imaginary and complex numbers. In the next few examples, we will apply many of the rules we are confident using with real numbers to solve problems involving purely imaginary numbers.

Example 2: Working with Positive Powers of Imaginary Numbers

Simplify (2𝑖)(βˆ’2𝑖).

Answer

In solving problems like this, it can be helpful to consider each part individually. Beginning with the first part, we can apply the properties of indices, or the commutativity of multiplication, to rewrite it as follows: (2𝑖)=2𝑖.

Recalling that, by definition, 𝑖=βˆ’1, this reduces to (2𝑖)=βˆ’4.

In a similar way, we consider the second term. By applying the rules of indices or the commutativity of multiplication, we can rewrite (βˆ’2𝑖)=(βˆ’2)𝑖.

We can easily evaluate (βˆ’2). However, how do we deal with π‘–οŠ©? For some students, the first time they see 𝑖 raised to a power other than two, they are unsure how to handle it. However, we already have all the tools we need to deal with this: if we simply rewrite 𝑖=π‘–Γ—π‘–οŠ©οŠ¨, we can use our knowledge that 𝑖=βˆ’1 to discover that 𝑖=βˆ’π‘–οŠ©. Therefore, the second part reduces to (βˆ’2𝑖)=8𝑖.

Finally, we can multiply the two parts together, which gives us our final answer of (2𝑖)(βˆ’2𝑖)=(βˆ’4)Γ—8𝑖=βˆ’32𝑖.

In the previous example, we found that 𝑖=βˆ’π‘–οŠ©. This raises the question of what happens as we raise 𝑖 to higher powers. We already know that 𝑖=π‘–οŠ§, 𝑖=βˆ’1, and 𝑖=βˆ’π‘–οŠ©. So what is 𝑖οŠͺ? We can calculate this in an analogous way to how we calculated π‘–οŠ© by noting that 𝑖=𝑖=(βˆ’1)=1.οŠͺ

Raising this equation to the 𝑛th power, we get 𝑖=1.οŠͺ

Multiplying this equation by the powers of 𝑖 from one to three, we get the following identities: 𝑖=1,𝑖=𝑖,𝑖=βˆ’1,𝑖=βˆ’π‘–.οŠͺοŠͺοŠͺοŠͺ

We can also express these identities as the following cycle.

Example 3: Powers of 𝑖

Simplify 1𝑖οŠͺ.

Answer

Firstly, we want to simplify 𝑖οŠͺ. To do this, we express 45 in the form 4π‘Ž+𝑏, where 𝑏 is an integer between 0 and 3. This will enable us to apply our knowledge of the powers of 𝑖 to eliminate the exponent from the expression.

Since 45=4Γ—11+1, we can express 𝑖=𝑖οŠͺοŠͺΓ—οŠ§οŠ§οŠ°οŠ§. Now we can apply our knowledge of the powers of 𝑖, in particular that 𝑖=𝑖οŠͺ, to simplify the expression to get 𝑖=𝑖οŠͺ. Alternatively, we could have approached this as follows: 𝑖=𝑖.οŠͺοŠͺΓ—οŠ§οŠ§οŠ°οŠ§

By applying the rules of exponents, we can express this as 𝑖=𝑖×𝑖=𝑖×𝑖.οŠͺοŠͺΓ—οŠ§οŠ§οŠͺ

Since 𝑖=1οŠͺ, we have 𝑖=(1)×𝑖=1×𝑖=𝑖.οŠͺ

Hence 1𝑖=1𝑖.οŠͺ

At this point, a student new to complex numbers might get a little stuck. However, we should not forget the algebraic and arithmetic tools we already know. Recall that when we want to rationalize the denominator in an expression like 1√2, we can multiply both the numerator and the denominator by √2, which gives us 12√2. In a similar way, thinking of 𝑖 as βˆšβˆ’1, we can apply the same technique, which results in the following calculation: 1𝑖=1𝑖×𝑖𝑖=𝑖𝑖.

Since 𝑖=βˆ’1, we have 1𝑖=π‘–βˆ’1=βˆ’π‘–.

Hence, we have 1𝑖=βˆ’π‘–.οŠͺ

In the previous example, we learnt how to deal with 1𝑖. Given that we can express 1𝑖 in index form as π‘–οŠ±οŠ§, and this is equal to βˆ’π‘–, we might start to wonder whether negative powers of 𝑖 also follow a similar cycle and the same rules we found for positive powers. As it turns out, they do and we have the following fact.

Theorem: Integer Powers of the Imaginary Number 𝑖

For all integers 𝑛, the following rules are true: 𝑖=1,𝑖=𝑖,𝑖=βˆ’1,𝑖=βˆ’π‘–.οŠͺοŠͺοŠͺοŠͺ

We can express this in a cycle as shown.

We can now look at an example of applying these rules.

Example 4: Simplifying Integer Powers of 𝑖

Given that 𝑛 is an integer, simplify π‘–οŠ§οŠ¬οŠοŠ±οŠ©οŠ«.

Answer

To apply the rules of the powers of 𝑖, we need to first express 16π‘›βˆ’35 in the form π‘Žπ‘š+𝑏, where 𝑏 is an integer between 0 and 3. We note that 16=4Γ—4 and 35=4Γ—8+3. Hence, 16π‘›βˆ’35=4Γ—4π‘›βˆ’(4Γ—8+3), which we can rewrite as 16π‘›βˆ’35=4(4π‘›βˆ’8)βˆ’3.

This is nearly in the correct form. We wanted to ensure that 𝑏 was between 0 and positive 3; however, here it is βˆ’3. We can easily solve this by expressing βˆ’3=βˆ’4+1. Substituting this back in gives 16π‘›βˆ’35=4(4π‘›βˆ’8)βˆ’4+1=4(4π‘›βˆ’9)+1.

Now we can apply the rules for integer powers of 𝑖, in particular 𝑖=𝑖οŠͺ, to get 𝑖=𝑖=𝑖.οŠͺ(οŠͺ)

We finish by looking at one last example of arithmetic with complex numbers where we need to be careful when trying to apply the familiar rule of arithmetic.

Example 5: Arithmetic with Imaginary Numbers

Simplify βˆšβˆ’10Γ—βˆšβˆ’6.

Answer

We need to be careful here not to fall into the trap of assuming that βˆšπ‘Žπ‘=βˆšπ‘Žβˆšπ‘ is true for all numbers. It is certainly true for positive real numbers. However, it is not true for negative numbers as we will see. To avoid this trap, we need to first express these square roots in terms of 𝑖 as follows: βˆšβˆ’10=π‘–βˆš10 and βˆšβˆ’6=π‘–βˆš6.

Now we can multiply them together and simplify: βˆšβˆ’10Γ—βˆšβˆ’6=π‘–βˆš10Γ—π‘–βˆš6=π‘–βˆš60.

Expressing 60 as a product of prime factors, 60=2Γ—3Γ—5, we can see that √60=2√15. Substituting this in and using the fact that 𝑖=βˆ’1, we find βˆšβˆ’10Γ—βˆšβˆ’6=βˆ’2√15.

Had we tried to use the rule βˆšπ‘Žπ‘=βˆšπ‘Žβˆšπ‘, we would have incorrectly concluded that the answer is 2√15.

This last example highlights the fact that although almost all of the rules of algebra and arithmetic can be applied to complex numbers, we need to be careful when we are dealing with fractional powers and roots to first express the square root of a negative number in terms of 𝑖 before trying to manipulate it.

Key Points

  1. We can solve many problems involving imaginary and complex numbers applying the familiar rules of arithmetic and algebra.
  2. We need to be careful when working with noninteger powers of any number which is not both positive and purely real; some of the rules we are familiar with do not apply to negative or complex numbers in general. For example, βˆšπ‘Žπ‘=βˆšπ‘Žβˆšπ‘ is not true for arbitrary complex numbers. In particular, it is not true for two negative numbers.
  3. The integer powers of 𝑖 form a cycle: 𝑖=1,𝑖=𝑖,𝑖=βˆ’1,𝑖=βˆ’π‘–.οŠͺοŠͺοŠͺοŠͺ

Using these rules, we can simplify some calculations involving complex numbers.

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