Lesson Explainer: Dividing Complex Numbers | Nagwa Lesson Explainer: Dividing Complex Numbers | Nagwa

Lesson Explainer: Dividing Complex Numbers Mathematics • First Year of Secondary School

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In this explainer, we will learn how to perform division on complex numbers.

When a student first encounters complex numbers, expressions like 3βˆ’6𝑖1βˆ’5𝑖 can seem a little mysterious or, at least, it can seem difficult to understand how one might compute the result. This explainer will connect this idea to more familiar areas of mathematics and help you understand how to evaluate expressions like this. Before we deal with division of complex numbers in general, we will consider the two simpler cases of division by a real number and division by a purely imaginary number.

Example 1: Dividing a Complex Number by a Real Number

Given 𝑧=5+3𝑖, express 𝑧2 in the form π‘Ž+𝑏𝑖.

Answer

Substituting in the value of 𝑧, we have 𝑧2=5+3𝑖2.

We can distribute the 12 over the complex number to get 𝑧2=52+32𝑖.

In many ways, dividing a complex number by a real number is a rather trivial exercise. However, dividing a complex number by an imaginary number is not so trivial as the next example will demonstrate.

Example 2: Dividing a Complex Number by an Imaginary Number

Simplify 2+4𝑖𝑖.

Answer

To simplify this fraction, we need to somehow convert the denominator to a real number. This can be accomplished by using the fact that 𝑖=βˆ’1. Hence, if we multiply both the numerator and the denominator by 𝑖, we will get a real number in the denominator which will enable us to simplify the fraction. Hence, 2+4𝑖𝑖=2+4𝑖𝑖×𝑖𝑖=(2+4𝑖)𝑖𝑖.

Distributing over the parentheses in the numerator, we have 2+4𝑖𝑖=2𝑖+4𝑖𝑖.

Using 𝑖=βˆ’1, we get 2+4𝑖𝑖=βˆ’4+2π‘–βˆ’1=4βˆ’2𝑖.

The technique we used above can be generalized to help us understand how to divide any two complex numbers. The first thing we need to do is identify a complex number which when multiplied by the denominator gives a real number. Then, we can multiply both the numerator and the denominator by this number and simplify. The question is, given a complex number 𝑧, what number when multiplied by 𝑧 results in a real number? This is the point where we should recall the properties of the complex conjugate, in particular, that for a complex number 𝑧=π‘Ž+𝑏𝑖, 𝑧𝑧=π‘Ž+𝑏, which is a real number. Hence, by multiplying the numerator and the denominator by the complex conjugate of the denominator, we can eliminate the imaginary part from the denominator and then simplify the result. This technique should not be new to most people. When learning about radicals, we face a similar problem trying to simplify expressions of the form π‘Ž+π‘βˆšπ‘π‘‘+π‘’βˆšπ‘“.

In this case, we multiply the numerator and the denominator by the conjugate of the denominator. This technique is often called rationalizing the denominator. With complex numbers, in many ways we are using the same technique in the special case where 𝑓 is a negative number.

Now, let’s consider an example where we have to simplify the division of two complex numbers, in a similar way to how we rationalize the denominator with radicals.

Example 3: Dividing Complex Numbers

Simplify 3βˆ’6𝑖1βˆ’5𝑖.

Answer

We begin by identifying a complex number that when multiplied by the denominator results in a real number. We usually use the complex conjugate of the denominator: 1+5𝑖. Now we multiply both the numerator and the denominator by this number as follows: 3βˆ’6𝑖1βˆ’5𝑖=3βˆ’6𝑖1βˆ’5𝑖×1+5𝑖1+5𝑖=(3βˆ’6𝑖)(1+5𝑖)(1βˆ’5𝑖)(1+5𝑖).

Expanding the parenthesis in the numerator and the denominator, we have 3βˆ’6𝑖1βˆ’5𝑖=3+15π‘–βˆ’6π‘–βˆ’30𝑖1+5π‘–βˆ’5π‘–βˆ’25𝑖.

Using 𝑖=βˆ’1 and gathering like terms, we have 3βˆ’6𝑖1βˆ’5𝑖=33+9𝑖26.

Finally, we express it in the form π‘Ž+𝑏𝑖 as follows: 3βˆ’6𝑖1βˆ’5𝑖=3326+926𝑖.

How To: Dividing Complex Numbers

To divide complex numbers, we use the following technique (sometimes referred to as β€œrealizing” the denominator):

  1. Multiply the numerator and denominator by the complex conjugate of the denominator.
  2. Expand the parenthesis in the numerator and denominator.
  3. Gather like terms (real and imaginary) remembering that 𝑖=βˆ’1.
  4. Express the answer in the form π‘Ž+𝑏𝑖 reducing any fractions.

Using this technique, we can actually derive a general form for the division of complex numbers as the next example will demonstrate.

Example 4: General Form of Complex Division

  1. Expand and simplify (𝑝+π‘žπ‘–)(π‘βˆ’π‘žπ‘–).
  2. Expand (π‘Ž+𝑏𝑖)(π‘βˆ’π‘žπ‘–).
  3. Hence, find a fraction which is equivalent to π‘Ž+𝑏𝑖𝑝+π‘žπ‘– and whose denominator is real.

Answer

Part 1

Expanding the parenthesis using FOIL or any other technique, we have (𝑝+π‘žπ‘–)(π‘βˆ’π‘žπ‘–)=π‘βˆ’π‘π‘žπ‘–+π‘π‘žπ‘–βˆ’π‘žπ‘–.

Using 𝑖=βˆ’1 and simplifying, we have (𝑝+π‘žπ‘–)(π‘βˆ’π‘žπ‘–)=𝑝+π‘ž.

Part 2

Similarly, we expand the parenthesis to get (π‘Ž+𝑏𝑖)(π‘βˆ’π‘žπ‘–)=π‘Žπ‘βˆ’π‘Žπ‘žπ‘–+π‘π‘π‘–βˆ’π‘π‘žπ‘–.

Gathering like terms and using 𝑖=βˆ’1, we have (π‘Ž+𝑏𝑖)(π‘βˆ’π‘žπ‘–)=(π‘Žπ‘+π‘π‘ž)+(π‘π‘βˆ’π‘Žπ‘ž)𝑖.

Part 3

To express this fraction with a real denominator, we multiply the numerator and the denominator by the complex conjugate of the denominator as follows: π‘Ž+𝑏𝑖𝑝+π‘žπ‘–=(π‘Ž+𝑏𝑖)(π‘βˆ’π‘žπ‘–)(𝑝+π‘žπ‘–)(π‘βˆ’π‘žπ‘–).

Substituting in our answers from part 1 and part 2, we have π‘Ž+𝑏𝑖𝑝+π‘žπ‘–=(π‘Žπ‘+π‘π‘ž)+(π‘π‘βˆ’π‘Žπ‘ž)𝑖𝑝+π‘ž.

Even though we have derived a general formula for complex division, it is preferable to be familiar with the technique rather than simply memorize the formula.

Example 5: Properties of Complex Division

If π‘Ž+𝑏𝑖=βˆ’3βˆ’5π‘–βˆ’3+5𝑖, is it true that π‘Ž+𝑏=1?

Answer

To express βˆ’3βˆ’5π‘–βˆ’3+5𝑖 in the form π‘Ž+𝑏𝑖, we multiply the numerator and the denominator by the complex conjugate of the denominator as follows: βˆ’3βˆ’5π‘–βˆ’3+5𝑖=(βˆ’3βˆ’5𝑖)(βˆ’3βˆ’5𝑖)(βˆ’3+5𝑖)(βˆ’3βˆ’5𝑖).

Expanding the parenthesis, we have βˆ’3βˆ’5π‘–βˆ’3+5𝑖=9+15𝑖+15𝑖+25𝑖9+15π‘–βˆ’15π‘–βˆ’25𝑖.

Using 𝑖=βˆ’1 and gathering like terms, we have βˆ’3βˆ’5π‘–βˆ’3+5𝑖=βˆ’16+30𝑖34.

Simplifying, we have βˆ’3βˆ’5π‘–βˆ’3+5𝑖=βˆ’817+1517𝑖.

Hence, π‘Ž=βˆ’817 and 𝑏=1517. Now we can consider the sum of their squares: π‘Ž+𝑏=ο€Όβˆ’817+ο€Ό1517=8+1517.

Simplifying gives π‘Ž+𝑏=64+225289=289289=1.

Hence, it is true that π‘Ž+𝑏=1.

The fact that π‘Ž+𝑏=1 in the previous question is no accident. In fact, this is an example of a general rule that if π‘Ž+𝑏𝑖=𝑧𝑧, for some complex number 𝑧, then π‘Ž+𝑏=1. This can be proved by working through the algebra. However, this is not very enlightening. Instead, results like this are best understood once we learn about the modulus and argument.

Example 6: Solving Complex Division Equations

Solve the equation 𝑧(2+𝑖)=3βˆ’π‘– for 𝑧.

Answer

We begin by dividing both sides of the equation by 2+𝑖 which results in the following equation: 𝑧=3βˆ’π‘–2+𝑖.

We now simplify the fraction by performing complex division. Hence, multiplying both the numerator and the denominator by the complex conjugate of the denominator, we get 𝑧=(3βˆ’π‘–)(2βˆ’π‘–)(2+𝑖)(2βˆ’π‘–).

Expanding the parenthesis, we have 𝑧=6βˆ’3π‘–βˆ’2𝑖+𝑖4+2π‘–βˆ’2π‘–βˆ’π‘–.

Using 𝑖=βˆ’1 and gathering like terms, we can rewrite this as 𝑧=5βˆ’5𝑖5=1βˆ’π‘–.

Due to the fact that multiplying and dividing complex numbers in this way can be fairly time consuming, it is useful to consider which approach will be the most efficient. This often involves using properties of complex numbers or noticing factors that we can quickly cancel. The next two examples will demonstrate how we can simplify our calculations.

Example 7: Complex Division

Simplify (βˆ’3+2𝑖)(3+3𝑖)(4+𝑖)(4+4𝑖).

Answer

When presented with an expression like this, it is good to first consider what approach we should take to solving it. We could expand the parenthesis in the numerator and the denominator and then multiply both the numerator and the denominator by the complex conjugate of the denominator. Alternatively, we could split the fraction in two and try to simplify each part then multiply the resulting complex numbers. The approach taken will generally depend on the particular expression given; however, it is good to look for features of the expression that might simplify the calculation. In this case, it is good to notice that we have a common factor of (1+𝑖) in both the numerator and the denominator. By canceling this factor first, we can simplify our calculation. Hence, (βˆ’3+2𝑖)(3+3𝑖)(4+𝑖)(4+4𝑖)=3(βˆ’3+2𝑖)(1+𝑖)4(4+𝑖)(1+𝑖)=34(βˆ’3+2𝑖)(4+𝑖).

We can now multiply both the numerator and the denominator by the complex conjugate of the denominator as follows: 34(βˆ’3+2𝑖)(4+𝑖)=34(βˆ’3+2𝑖)(4βˆ’π‘–)(4+𝑖)(4βˆ’π‘–).

Expanding the parentheses in the numerator and the denominator, we have 34(βˆ’3+2𝑖)(4βˆ’π‘–)(4+𝑖)(4βˆ’π‘–)=34(βˆ’12+3𝑖+8π‘–βˆ’2𝑖)(16+4π‘–βˆ’4π‘–βˆ’π‘–).

Using 𝑖=βˆ’1 and gathering like terms, we can rewrite this as 34ο€Ήβˆ’12+3𝑖+8π‘–βˆ’2𝑖(16+4π‘–βˆ’4π‘–βˆ’π‘–)=34(βˆ’10+11𝑖)17.

Finally, we can simplify to get 34ο€Ήβˆ’12+3𝑖+8π‘–βˆ’2𝑖(16+4π‘–βˆ’4π‘–βˆ’π‘–)=βˆ’1534+3368𝑖.

For the next question, we will again look at an example where applying the properties of complex numbers can simplify out calculations.

Example 8: Complex Expressions Involving Division

Simplify 3βˆ’4𝑖2+2𝑖+3βˆ’4𝑖2βˆ’2𝑖.

Answer

It is possible to solve this problem by performing the complex division on both of the fractions and then adding their results. However, we can simplify our calculation by first noticing that we can factor out 3βˆ’4𝑖 from both terms. Hence, we can rewrite the expression as

3βˆ’4𝑖2+2𝑖+3βˆ’4𝑖2βˆ’2𝑖=(3βˆ’4𝑖)ο€Ό12+2𝑖+12βˆ’2π‘–οˆ.(1)

Now we consider the expression in the parentheses; notice that the denominators of the two fractions are a complex conjugate pair; that is, the expression is in the form 1𝑧+1𝑧.

If we express this as a single fraction over a common denominator, we have 1𝑧+1𝑧=𝑧+𝑧𝑧𝑧.

Using the properties of complex conjugates, we know that if 𝑧=π‘Ž+𝑏𝑖, 𝑧𝑧=π‘Ž+π‘οŠ¨οŠ¨ and 𝑧+𝑧=2(𝑧)=2π‘ŽRe. Hence, 1𝑧+1𝑧=2π‘Žπ‘Ž+𝑏.

Therefore, ο€Ό12+2𝑖+12βˆ’2π‘–οˆ=2Γ—22+2=12.

Substituting this into (1), we have 3βˆ’4𝑖2+2𝑖+3βˆ’4𝑖2βˆ’2𝑖=12(3βˆ’4𝑖)=32βˆ’2𝑖.

Finally, let’s consider an example where we have to find missing values in an equation by dividing complex numbers.

Example 9: Solving a Two-Variable Linear Equation with Complex Coefficients

Given that π‘₯+𝑖𝑦1βˆ’3π‘–βˆ’9π‘₯βˆ’π‘–π‘¦1+3𝑖=6+5𝑖4βˆ’8𝑖, where π‘₯ and 𝑦 are real numbers, determine the value of π‘₯ and the value of 𝑦.

Answer

In this example, we want to determine the missing values π‘₯ and 𝑦 in a two-variable linear equation with complex coefficients.

The given equation contains 3 separate complex divisions, two on the left hand side and one on the right hand side of the equation. We begin by simplifying each term by performing the complex division. This is achieved by multiplying the numerator and denominator by the complex conjugate of the denominator, which results in a real number in the denominator after distributing over the parentheses.

For the first term, π‘₯+𝑖𝑦1βˆ’3𝑖, we multiply the denominator and numerator by 1+3𝑖, which is the complex conjugate of the denominator: π‘₯+𝑖𝑦1βˆ’3𝑖=π‘₯+𝑖𝑦1βˆ’3𝑖×1+3𝑖1+3𝑖=(π‘₯+𝑖𝑦)(1+3𝑖)(1βˆ’3𝑖)(1+3𝑖).

Distributing over the parentheses in the numerator and denominator, we have π‘₯+𝑖𝑦1βˆ’3𝑖=π‘₯+3π‘₯𝑖+𝑦𝑖+3𝑖𝑦1+3π‘–βˆ’3π‘–βˆ’9𝑖=π‘₯+3π‘₯𝑖+π‘¦π‘–βˆ’3𝑦1+9=(π‘₯βˆ’3𝑦)+𝑖(3π‘₯+𝑦)10.

Repeating this process for the second term, 9π‘₯βˆ’π‘–π‘¦1+3𝑖, this time multiplying the numerator and denominator by 1βˆ’3𝑖, we obtain 9π‘₯βˆ’π‘–π‘¦1+3𝑖=9π‘₯βˆ’π‘–π‘¦1+3𝑖×1βˆ’3𝑖1βˆ’3𝑖=(9π‘₯βˆ’π‘–π‘¦)(1βˆ’3𝑖)(1+3𝑖)(1βˆ’3𝑖)=9π‘₯βˆ’27π‘₯π‘–βˆ’π‘¦π‘–+3𝑖𝑦1+3π‘–βˆ’3π‘–βˆ’9𝑖=9π‘₯βˆ’27π‘₯π‘–βˆ’π‘¦π‘–βˆ’3𝑦1+9=(9π‘₯βˆ’3𝑦)+𝑖(βˆ’27π‘₯βˆ’π‘¦)10.

Finally, for the last term, 6+5𝑖4βˆ’8𝑖, on the right-hand side, by multiplying the denominator and numerator by 4+8𝑖, 6+5𝑖4βˆ’8𝑖=6+5𝑖4βˆ’8𝑖×4+8𝑖4+8𝑖=(6+5𝑖)(4+8𝑖)(4βˆ’8𝑖)(4+8𝑖)=24+48𝑖+20𝑖+40𝑖16+32π‘–βˆ’32π‘–βˆ’64𝑖=24+48𝑖+20π‘–βˆ’4016+64=βˆ’16+68𝑖80.

Thus, the given equation becomes π‘₯+𝑖𝑦1βˆ’3π‘–βˆ’9π‘₯βˆ’π‘–π‘¦1+3𝑖=6+5𝑖4βˆ’8𝑖(π‘₯βˆ’3𝑦)+𝑖(3π‘₯+𝑦)10βˆ’(9π‘₯βˆ’3𝑦)+𝑖(βˆ’27π‘₯βˆ’π‘¦)10=βˆ’16+68𝑖80(π‘₯βˆ’3𝑦)βˆ’(9π‘₯βˆ’3𝑦)+𝑖(3π‘₯+𝑦)+𝑖(27π‘₯+𝑦)10=βˆ’16+68𝑖80βˆ’8π‘₯+𝑖(30π‘₯+2𝑦)10=βˆ’16+68𝑖80βˆ’8π‘₯+𝑖(30π‘₯+2𝑦)=βˆ’16+68𝑖8.

The real values of π‘₯ and 𝑦 can be found by equating the real and imaginary parts of both sides. First equating the real parts, we have βˆ’8π‘₯=βˆ’168π‘₯=14.

Now, the imaginary parts, 30π‘₯+2𝑦=688𝑦=6816βˆ’15π‘₯=6816βˆ’154=12.

To summarize, the real solutions to the given equation are π‘₯=14,𝑦=12.

Let’s summarize some of the key points that we covered in this explainer.

Key Points

  • Dividing complex numbers uses the same technique as for rationalizing the denominator.
  • To divide complex numbers, we multiply the numerator and the denominator by the complex conjugate of the denominator, and then we expand the brackets and simplify using 𝑖=βˆ’1.
  • In expressions involving multiplication and division of multiple complex numbers, it is useful to look for common factors or whether we can apply some of the properties of complex numbers to simplify our calculations.

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