Multiply and divide complex numbers

Peter Sallay, San Jacinto College

110 Multiply and divide complex numbers

Multiplying Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Example 4: Multiplying a Complex Number by a Real Number

Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i. — **Figure 5**

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

How To: Given a complex number and a real number, multiply to find the product.

Use the distributive property.
Simplify.

Example 5: Multiplying a Complex Number by a Real Number

Find the product $4 (2 + 5 i)$ .

Solution

Distribute the 4.

{\begin{cases} 4 (2 + 5 i) = (4 \cdot 2) + (4 \cdot 5 i) \\ = 8 + 20 i \end{cases}

Try It 4

Find the product $- 4 (2 + 6 i)$ .

Solution

Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

(a + b i) (c + d i) = a c + a d i + b c i + b d i^{2}

Because $i^{2} = - 1$ , we have

(a + b i) (c + d i) = a c + a d i + b c i - b d

To simplify, we combine the real parts, and we combine the imaginary parts.

(a + b i) (c + d i) = (a c - b d) + (a d + b c) i

How To: Given two complex numbers, multiply to find the product.

Use the distributive property or the FOIL method.
Simplify.

Example 6: Multiplying a Complex Number by a Complex Number

Multiply $(4 + 3 i) (2 - 5 i)$ .

Solution

Use $(a + b i) (c + d i) = (a c - b d) + (a d + b c) i$

{\begin{cases} (4 + 3 i) (2 - 5 i) = (4 \cdot 2 - 3 \cdot (- 5)) + (4 \cdot (- 5) + 3 \cdot 2) i \\ = (8 + 15) + (- 20 + 6) i \\ = 23 - 14 i \end{cases}

Try It 5

Multiply $(3 - 4 i) (2 + 3 i)$ .

Solution

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of $a + b i$ is $a - b i$ .

Note that complex conjugates have a reciprocal relationship: The complex conjugate of $a + b i$ is $a - b i$ , and the complex conjugate of $a - b i$ is $a + b i$ . Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide $c + d i$ by $a + b i$ , where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

\frac{c + d i}{a + b i} where a \neq 0 and b \neq 0

Multiply the numerator and denominator by the complex conjugate of the denominator.

\frac{(c + d i)}{(a + b i)} \cdot \frac{(a - b i)}{(a - b i)} = \frac{(c + d i) (a - b i)}{(a + b i) (a - b i)}

Apply the distributive property.

= \frac{c a - c b i + a d i - b d i^{2}}{a^{2} - a b i + a b i - b^{2} i^{2}}

Simplify, remembering that $i^{2} = - 1$ .

{\begin{cases} = \frac{c a - c b i + a d i - b d (- 1)}{a^{2} - a b i + a b i - b^{2} (- 1)} \\ = \frac{(c a + b d) + (a d - c b) i}{a^{2} + b^{2}} \end{cases}

A General Note: The Complex Conjugate

The complex conjugate of a complex number $a + b i$ is $a - b i$ . It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

When a complex number is multiplied by its complex conjugate, the result is a real number.
When a complex number is added to its complex conjugate, the result is a real number.

Example 7: Finding Complex Conjugates

Find the complex conjugate of each number.

$2 + i \sqrt{5}$
$- \frac{1}{2} i$

Solution

The number is already in the form $a + b i / /$ . The complex conjugate is $a - b i$ , or $2 - i \sqrt{5}$ .
We can rewrite this number in the form $a + b i$ as $0 - \frac{1}{2} i$ . The complex conjugate is $a - b i$ , or $0 + \frac{1}{2} i$ . This can be written simply as $\frac{1}{2} i$ .

Analysis of the Solution

Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.

How To: Given two complex numbers, divide one by the other.

Write the division problem as a fraction.
Determine the complex conjugate of the denominator.
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
Simplify.

Example 8: Dividing Complex Numbers

Divide $(2 + 5 i)$ by $(4 - i)$ .

Solution

We begin by writing the problem as a fraction.

\frac{(2 + 5 i)}{(4 - i)}

Then we multiply the numerator and denominator by the complex conjugate of the denominator.

\frac{(2 + 5 i)}{(4 - i)} \cdot \frac{(4 + i)}{(4 + i)}

To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).

{\begin{cases} \frac{(2 + 5 i)}{(4 - i)} \cdot \frac{(4 + i)}{(4 + i)} = \frac{8 + 2 i + 20 i + 5 i^{2}}{16 + 4 i - 4 i - i^{2}} \\ = \frac{8 + 2 i + 20 i + 5 (- 1)}{16 + 4 i - 4 i - (- 1)} & Because i^{2} = - 1 \\ = \frac{3 + 22 i}{17} \\ = \frac{3}{17} + \frac{22}{17} i & Separate real and imaginary parts . \end{cases}

Note that this expresses the quotient in standard form.

Example 9: Substituting a Complex Number into a Polynomial Function

Let $f (x) = x^{2} - 5 x + 2$ . Evaluate $f (3 + i)$ .

Solution

Substitute 3 plus i for x. Multiply. Substitute negative 1 for i squared. Combine like terms. — **Figure 6**

Substitute $x = 3 + i$ into the function $f (x) = x^{2} - 5 x + 2$ and simplify.

Analysis of the Solution

We write $f (3 + i) = - 5 + i$ . Notice that the input is $3 + i$ and the output is $- 5 + i$ .

Try It 6

Let $f (x) = 2 x^{2} - 3 x$ . Evaluate $f (8 - i)$ .

Solution

Example 10: Substituting an Imaginary Number in a Rational Function

Let $f (x) = \frac{2 + x}{x + 3}$ . Evaluate $f (10 i)$ .

Solution

Substitute $x = 10 i$ and simplify.

{\begin{cases} \frac{2 + 10 i}{10 i + 3} & Substitute 10 i for x . \\ \frac{2 + 10 i}{3 + 10 i} & Rewrite the denominator in standard form . \\ \frac{2 + 10 i}{3 + 10 i} \cdot \frac{3 - 10 i}{3 - 10 i} & Prepare to multiply the numerator and \\ denominator by the complex conjugate \\ of the denominator . \\ \frac{6 - 20 i + 30 i - 100 i^{2}}{9 - 30 i + 30 i - 100 i^{2}} & Multiply using the distributive property or the FOIL method . \\ \frac{6 - 20 i + 30 i - 100 (- 1)}{9 - 30 i + 30 i - 100 (- 1)} & Substitute - 1 for i^{2} . \\ \frac{106 + 10 i}{109} & Simplify . \\ \frac{106}{109} + \frac{10}{109} i & Separate the real and imaginary parts . \end{cases}

Try It 7

Let $f (x) = \frac{x + 1}{x - 4}$ . Evaluate $f (- i)$ .

Solution

Simplifying Powers of i

The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.

{\begin{cases} i^{1} = i \\ i^{2} = - 1 \\ i^{3} = i^{2} \cdot i = - 1 \cdot i = - i \\ i^{4} = i^{3} \cdot i = - i \cdot i = - i^{2} = - (- 1) = 1 \\ i^{5} = i^{4} \cdot i = 1 \cdot i = i \end{cases}

We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.

{\begin{cases} i^{6} = i^{5} \cdot i = i \cdot i = i^{2} = - 1 \\ i^{7} = i^{6} \cdot i = i^{2} \cdot i = i^{3} = - i \\ i^{8} = i^{7} \cdot i = i^{3} \cdot i = i^{4} = 1 \\ i^{9} = i^{8} \cdot i = i^{4} \cdot i = i^{5} = i \end{cases}

Example 11: Simplifying Powers of i

Evaluate $i^{35}$ .

Solution

Since $i^{4} = 1$ , we can simplify the problem by factoring out as many factors of $i^{4}$ as possible. To do so, first determine how many times 4 goes into 35: $35 = 4 \cdot 8 + 3$ .

i^{35} = i^{4 \cdot 8 + 3} = i^{4 \cdot 8} \cdot i^{3} = {(i^{4})}^{8} \cdot i^{3} = 1^{8} \cdot i^{3} = i^{3} = - i

Q & A

Can we write $i^{35}$ in other helpful ways?

As we saw in Example 11, we reduced $i^{35}$ to $i^{3}$ by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of $i^{35}$ may be more useful. The table below shows some other possible factorizations.

Factorization of $i^{35}$	$i^{34} \cdot i$	$i^{33} \cdot i^{2}$	$i^{31} \cdot i^{4}$	$i^{19} \cdot i^{16}$
Reduced form	${(i^{2})}^{17} \cdot i$	$i^{33} \cdot (- 1)$	$i^{31} \cdot 1$	$i^{19} \cdot {(i^{4})}^{4}$
Simplified form	${(- 1)}^{17} \cdot i$	$- i^{33}$	$i^{31}$	$i^{19}$

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

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Multiplying Complex Numbers

Example 4: Multiplying a Complex Number by a Real Number

How To: Given a complex number and a real number, multiply to find the product.

Example 5: Multiplying a Complex Number by a Real Number

Solution

Try It 4

Multiplying Complex Numbers Together

How To: Given two complex numbers, multiply to find the product.

Example 6: Multiplying a Complex Number by a Complex Number

Solution

Try It 5

Dividing Complex Numbers

A General Note: The Complex Conjugate

Example 7: Finding Complex Conjugates

Solution

Analysis of the Solution

How To: Given two complex numbers, divide one by the other.

Example 8: Dividing Complex Numbers

Solution

Example 9: Substituting a Complex Number into a Polynomial Function

Solution

Analysis of the Solution

Try It 6

Example 10: Substituting an Imaginary Number in a Rational Function

Solution

Try It 7

Simplifying Powers of i

Example 11: Simplifying Powers of i

Solution

Q & A

License

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