Lesson Explainer: Equating, Adding, and Subtracting Complex Numbers | Nagwa Lesson Explainer: Equating, Adding, and Subtracting Complex Numbers | Nagwa

Lesson Explainer: Equating, Adding, and Subtracting Complex Numbers Mathematics

In this explainer, we will learn how to equate, add, and subtract complex numbers.

We begin by recalling the definition of a complex number and some notations for complex numbers.

Definition: Complex Numbers

A complex number is a number of the form 𝑎+𝑏𝑖, where both 𝑎 and 𝑏 are real numbers and 𝑖 is the square root of 1. The set of all complex numbers is denoted by .

For a complex number 𝑧=𝑎+𝑏𝑖, we define the real part of 𝑧 to be 𝑎 and write Re(𝑧)=𝑎.

Similarly, we define the imaginary part of 𝑧 to be 𝑏 and write Im(𝑧)=𝑏.

Some books and articles use the notation (𝑧) and (𝑧) to refer to the real and imaginary parts of 𝑧.

Before we start doing arithmetic with complex numbers, we need to understand what it means for two complex numbers to be equal. We define equality of complex numbers in a similar way to how we define equality of algebraic expressions involving variables. For example, if 𝑥 is a variable and 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers, saying the two algebraic expressions 𝑎+𝑏𝑥 and 𝑐+𝑑𝑥 are equal is equivalent to stating that 𝑎=𝑐 and 𝑏=𝑑. We define the equality of complex numbers in a similar way.

Definition: Equality of Complex Numbers

Two complex numbers 𝑧=𝑎+𝑏𝑖 and 𝑧=𝑐+𝑑𝑖 are said to be equal if 𝑎=𝑐 and 𝑏=𝑑. Conversely, if 𝑧=𝑧, then 𝑎=𝑐 and 𝑏=𝑑. Equivalently, we can state that two complex numbers 𝑧 and 𝑧 are equal if ReRe(𝑧)=(𝑧) and ImIm(𝑧)=(𝑧) and the equivalent converse statement.

Oftentimes, working with the second version of this is easiest as we will see in some of the examples below.

Let us begin with an example where we can apply the equality of complex numbers to identify unknown values in given complex expressions.

Example 1: Equality of Complex Numbers

If the complex numbers 7+𝑎𝑖 and 𝑏3𝑖 are equal, what are the values of 𝑎 and 𝑏?

Answer

Recall that two complex numbers are said to be equal if both their real and imaginary parts are equal. Beginning with equating the real parts, we have 7=𝑏. Similarly, equating the imaginary parts, we arrive at the equation 𝑎=3 (we need to be careful not to miss the negative sign here). Hence, we have 𝑎=3 and 𝑏=7.

In the previous example, we were able to identify two different unknown constants by using one complex equality. Since the equality for complex numbers equates the real and the imaginary parts separately, one complex equality generates two separate equations. This allows us to identify two different unknowns from one complex equality. Let us consider another example where we find two unknowns from an equation involving complex numbers.

Example 2: Solving Simple Equations Involving Complex Numbers

Determine the real numbers 𝑥 and 𝑦 that satisfy the equation 5𝑥+2+(3𝑦5)𝑖=3+4𝑖.

Answer

Recall that two complex numbers are said to be equal if both their real and imaginary parts are equal. By considering the real and imaginary parts separately, we can derive two equations which we can then solve for 𝑥 and 𝑦. Since we are told that 𝑥 and 𝑦 are real numbers, we know that 5𝑥+2 and 3𝑦5 are, respectively, the real and imaginary parts of the complex number on the left-hand side of the given equation. Starting with the real parts, we have 5𝑥+2=3.

Subtracting 2 from both sides gives 5𝑥=5.

Then, by dividing by 5, we get 𝑥=1.

Taking the imaginary parts of both sides, we have 3𝑦5=4.

Adding 5 to both sides gives 3𝑦=9; then dividing by 3 gives 𝑦=3.

Hence, 𝑥=1 and 𝑦=3.

Similar to how we defined the equality of complex numbers, the basic tenets of addition and subtraction of complex numbers are analogous to their equivalents within the algebra of polynomials. To add and subtract polynomials, we add and subtract the corresponding coefficients.

Definition: Addition and Subtraction of Complex Numbers

For two complex numbers 𝑧=𝑎+𝑏𝑖 and 𝑧=𝑐+𝑑𝑖, we define 𝑧+𝑧=(𝑎+𝑐)+(𝑏+𝑑)𝑖.

Similarly,𝑧𝑧=(𝑎𝑐)+(𝑏𝑑)𝑖.

In other words, we add or subtract complex numbers by separately adding or subtracting their real parts and imaginary parts.

Alternatively, we can add or subtract complex numbers by expanding through the parenthesis and gathering real and imaginary terms. Using this method, 𝑧+𝑧=(𝑎+𝑏𝑖)+(𝑐+𝑑𝑖)=𝑎+𝑏𝑖+𝑐+𝑑𝑖=(𝑎+𝑐)+(𝑏+𝑑)𝑖.

We note that this leads to the same expression as above. To subtract two complex numbers: 𝑧𝑧=(𝑎+𝑏𝑖)(𝑐+𝑑𝑖), where we need to be careful to expand (𝑐+𝑑𝑖)=𝑐𝑑𝑖. This leads to 𝑎+𝑏𝑖𝑐𝑑𝑖=(𝑎𝑐)+(𝑏𝑑)𝑖, which is also the same as above.

This statement can be generalized to encompass the situation when we add or subtract multiple complex numbers. In such cases, we separately add or subtract the real part and imaginary part of each complex number. In our next example, we will add and subtract multiple complex numbers.

Example 3: Adding and Subtracting Complex Numbers

What is 9+(7+4𝑖)+(44𝑖)(1+3𝑖)?

Answer

We will demonstrate two different methods here.

Method 1: Recall that we can add or subtract multiple complex numbers by adding or subtracting the real part and the imaginary part of each complex number separately. Starting with the real parts, we have 9+7+(4)1=7.

So the real part of the result is 7. Similarly, for the imaginary parts, we have 4+(4)3=3.

Putting these two parts together, we have that the result is 73𝑖.

Method 2: Alternatively, we can add or subtract complex numbers by expanding through the parenthesis and gathering real and imaginary terms. In particular, we should be careful when expanding the last parenthesis since there is a negative sign in front of the parenthesis. We know that (1+3𝑖)=13𝑖. Using this method, we can write 9+(7+4𝑖)+(44𝑖)(1+3𝑖)=9+7+4𝑖44𝑖13𝑖=(9+741)+(443)𝑖, which simplifies to 73𝑖.

In practice, we often use the method of gathering like terms. However, occasionally, it can be useful to remember that there is an alternative method as we will see in the example below.

Example 4: Subtracting Complex Numbers

If 𝑟=5+2𝑖 and 𝑠=9𝑖, find Re(𝑟𝑠).

Answer

We will present two different methods for this example, but we will observe that the second method is preferable since it is much simpler.

Method 1: We begin by calculating 𝑟𝑠 by gathering like terms. Firstly, substituting the values of 𝑟 and 𝑠, we have 𝑟𝑠=5+2𝑖(9𝑖).

At this point, we need to be careful with the minus signs. Multiplying each term in the bracket by 1 gives 𝑟𝑠=5+2𝑖9+𝑖.

By gathering like terms, this simplifies to 𝑟𝑠=(59)+(2+1)𝑖=4+3𝑖.

Taking the real part, we have Re(𝑟𝑠)=4.

While this method does lead to a correct solution, it requires more calculations than necessary. Specifically, we have unnecessarily calculated the imaginary part of 𝑟𝑠.

Method 2: We remember that the real part of the difference of two complex numbers is the difference of their real parts: ReReRe(𝑟𝑠)=(𝑟)(𝑠).

We can simplify our calculation as follows: ReReRe(𝑟𝑠)=(5+2𝑖)(9𝑖)=59=4.

We finish by looking at one slightly more advanced example.

Example 5: Solving Equations Involving Complex Numbers

Let 𝑧=4𝑥+2𝑦𝑖 and 𝑧=4𝑦+𝑥𝑖, where 𝑥, 𝑦. Given that 𝑧𝑧=5+2𝑖, find 𝑧 and 𝑧.

Answer

We can subtract complex numbers by subtracting the real and the imaginary parts separately. Also, equality in complex numbers implies both the equality of the real parts and the equality of the imaginary parts of the complex numbers. By considering the real and imaginary parts separately, we can derive two equations which we can then solve for 𝑥 and 𝑦. Starting with the real part, we have ReReRe(𝑧)(𝑧)=(5+2𝑖).

Since we are told that 𝑥 and 𝑦 are real numbers, we know that 4𝑥 and 4𝑦 are the real parts of 𝑧 and 𝑧 respectively. This tells us that ReRe(𝑧)(𝑧)=4𝑥4𝑦. Substituting this expression in the equation above, we obtain

4𝑥4𝑦=5.(1)

Similarly, by considering the imaginary parts, we have ImImIm(𝑧)(𝑧)=(5+2𝑖).

Since 2𝑦 and 𝑥 are the imaginary parts of 𝑧 and 𝑧, respectively, we have ImIm(𝑧)(𝑧)=2𝑦𝑥. Substituting this expression to the equation above, we obtain 2𝑦𝑥=2.

Rearranging to make 𝑥 the subject, we get

𝑥=2𝑦2.(2)

Substituting this into (1) gives 4(2𝑦2)4𝑦=5.

At this point, we could multiply out the brackets. However, it is more efficient to divide both sides of the equation by 4 as follows: 2𝑦2𝑦=54.

Adding two to both sides and simplifying gives 𝑦=134. Substituting this back into (2) gives 𝑥=21342.=92.

Having found 𝑥 and 𝑦, we might be tempted to stop. However, the question actually asked us for 𝑧 and 𝑧. Hence, we still need to substitute these values back into the equations for 𝑧 and 𝑧 to finish. Starting with 𝑧, we have 𝑧=492+2134𝑖=18+132𝑖.

Similarly, 𝑧=4134+92𝑖=13+92𝑖.

It is always a good practice to check your answer. Therefore, we subtract 𝑧 from 𝑧 and get 5+2𝑖 as expected.

Let us recap a few important concepts from this explainer.

Key Points

  • Addition, subtraction, and equality of complex numbers are defined in an analogous way to addition, subtraction, and equality of polynomial expressions.
  • By applying familiar rules of algebra, we can begin to work effectively with complex numbers.
  • For complex numbers 𝑧=𝑎+𝑏𝑖 and 𝑧=𝑐+𝑑𝑖,
    • 𝑧 = 𝑧 is equivalent to saying both their real and imaginary parts are equal or 𝑎=𝑐 and 𝑏=𝑑;
    • 𝑧±𝑧=(𝑎±𝑐)+(𝑏±𝑑)𝑖;
    • ReReRe(𝑧±𝑧)=(𝑧)±(𝑧);
    • ImImIm(𝑧±𝑧)=(𝑧)±(𝑧).

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