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Dividing complex numbers

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

We begin by writing the problem as a fraction.

( 2 + 5 i ) ( 4 i )

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

( 2 + 5 i ) ( 4 i ) ( 4 + i ) ( 4 + i )

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

( 2 + 5 i ) ( 4 i ) ( 4 + i ) ( 4 + i ) = 8 + 2 i + 20 i + 5 i 2 16 + 4 i 4 i i 2                             = 8 + 2 i + 20 i + 5 ( 1 ) 16 + 4 i 4 i ( 1 ) Because    i 2 = 1                             = 3 + 22 i 17                             = 3 17 + 22 17 i Separate real and imaginary parts .

Note that this expresses the quotient in standard form.

Substituting a complex number into a polynomial function

Let f ( x ) = x 2 5 x + 2. Evaluate f ( 3 + i ) .

Substitute x = 3 + i into the function f ( x ) = x 2 5 x + 2 and simplify.

Let f ( x ) = 2 x 2 3 x . Evaluate f ( 8 i ) .

102 29 i

Substituting an imaginary number in a rational function

Let f ( x ) = 2 + x x + 3 . Evaluate f ( 10 i ) .

Substitute x = 10 i and simplify.

2 + 10 i 10 i + 3 Substitute  10 i  for  x . 2 + 10 i 3 + 10 i Rewrite the denominator in standard form . 2 + 10 i 3 + 10 i 3 10 i 3 10 i Prepare to multiply the numerator and denominator by the complex conjugate of the denominator . 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 . 6 20 i + 30 i 100 ( 1 ) 9 30 i + 30 i 100 ( 1 ) Substitute –1 for   i 2 . 106 + 10 i 109 Simplify . 106 109 + 10 109 i Separate the real and imaginary parts .

Let f ( x ) = x + 1 x 4 . Evaluate f ( i ) .

3 17 + 5 i 17

Simplifying powers of i

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

i 1 = i i 2 = 1 i 3 = i 2 i = 1 i = i i 4 = i 3 i = i i = i 2 = ( 1 ) = 1 i 5 = i 4 i = 1 i = i

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 .

i 6 = i 5 i = i i = i 2 = 1 i 7 = i 6 i = i 2 i = i 3 = i i 8 = i 7 i = i 3 i = i 4 = 1 i 9 = i 8 i = i 4 i = i 5 = i

Simplifying powers of i

Evaluate i 35 .

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 8 + 3.

i 35 = i 4 8 + 3 = i 4 8 i 3 = ( i 4 ) 8 i 3 = 1 8 i 3 = i 3 = i

Can we write i 35 in other helpful ways?

As we saw in [link] , 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. [link] shows some other possible factorizations.

Factorization of i 35 i 34 i i 33 i 2 i 31 i 4 i 19 i 16
Reduced form ( i 2 ) 17 i i 33 ( 1 ) i 31 1 i 19 ( i 4 ) 4
Simplified form ( 1 ) 17 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.

Access these online resources for additional instruction and practice with complex numbers.

Key concepts

  • The square root of any negative number can be written as a multiple of i . See [link] .
  • To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See [link] .
  • Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See [link] .
  • Complex numbers can be multiplied and divided.
  • To multiply complex numbers, distribute just as with polynomials. See [link] , [link] , and [link] .
  • To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See [link] , [link] , and [link] .
  • The powers of i are cyclic, repeating every fourth one. See [link] .

Verbal

Explain how to add complex numbers.

Add the real parts together and the imaginary parts together.

What is the basic principle in multiplication of complex numbers?

Give an example to show the product of two imaginary numbers is not always imaginary.

i times i equals –1, which is not imaginary. (answers vary)

What is a characteristic of the plot of a real number in the complex plane?

Algebraic

For the following exercises, evaluate the algebraic expressions.

If  f ( x ) = x 2 + x 4 , evaluate f ( 2 i ) .

8 + 2 i

If  f ( x ) = x 3 2 , evaluate f ( i ) .

If  f ( x ) = x 2 + 3 x + 5 , evaluate f ( 2 + i ) .

14 + 7 i

If  f ( x ) = 2 x 2 + x 3 , evaluate f ( 2 3 i ) .

If  f ( x ) = x + 1 2 x , evaluate f ( 5 i ) .

23 29 + 15 29 i

If  f ( x ) = 1 + 2 x x + 3 , evaluate f ( 4 i ) .

Graphical

For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown.

Graph of a parabola intersecting the real axis.

2 real and 0 nonreal

Graph of a parabola not intersecting the real axis.

For the following exercises, plot the complex numbers on the complex plane.

1 2 i

Graph of the plotted point, 1-2i.

2 + 3 i

i

Graph of the plotted point, i.

3 4 i

Numeric

For the following exercises, perform the indicated operation and express the result as a simplified complex number.

( 3 + 2 i ) + ( 5 3 i )

8 i

( 2 4 i ) + ( 1 + 6 i )

( 5 + 3 i ) ( 6 i )

11 + 4 i

( 2 3 i ) ( 3 + 2 i )

( 4 + 4 i ) ( 6 + 9 i )

2 5 i

( 2 + 3 i ) ( 4 i )

( 5 2 i ) ( 3 i )

6 + 15 i

( 6 2 i ) ( 5 )

( 2 + 4 i ) ( 8 )

16 + 32 i

( 2 + 3 i ) ( 4 i )

( 1 + 2 i ) ( 2 + 3 i )

4 7 i

( 4 2 i ) ( 4 + 2 i )

( 3 + 4 i ) ( 3 4 i )

25

3 + 4 i 2

6 2 i 3

2 2 3 i

5 + 3 i 2 i

6 + 4 i i

4 6 i

2 3 i 4 + 3 i

3 + 4 i 2 i

2 5 + 11 5 i

2 + 3 i 2 3 i

9 + 3 16

15 i

4 4 25

2 + 12 2

1 + i 3

4 + 20 2

i 8

1

i 15

i 22

1

Technology

For the following exercises, use a calculator to help answer the questions.

Evaluate ( 1 + i ) k for k = 4, 8, and 12 . Predict the value if k = 16.

Evaluate ( 1 i ) k for k = 2, 6, and 10 . Predict the value if k = 14.

128i

Evaluate ( 1 + i ) k ( 1 i ) k for k = 4, 8, and 12 . Predict the value for k = 16.

Show that a solution of x 6 + 1 = 0 is 3 2 + 1 2 i .

( 3 2 + 1 2 i ) 6 = 1

Show that a solution of x 8 1 = 0 is 2 2 + 2 2 i .

Extensions

For the following exercises, evaluate the expressions, writing the result as a simplified complex number.

1 i + 4 i 3

3 i

1 i 11 1 i 21

i 7 ( 1 + i 2 )

0

i −3 + 5 i 7

( 2 + i ) ( 4 2 i ) ( 1 + i )

5 – 5i

( 1 + 3 i ) ( 2 4 i ) ( 1 + 2 i )

( 3 + i ) 2 ( 1 + 2 i ) 2

2 i

3 + 2 i 2 + i + ( 4 + 3 i )

4 + i i + 3 4 i 1 i

9 2 9 2 i

3 + 2 i 1 + 2 i 2 3 i 3 + i

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Source:  OpenStax, Essential precalculus, part 1. OpenStax CNX. Aug 26, 2015 Download for free at http://legacy.cnx.org/content/col11871/1.1
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