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# Complex Numbers Help (page 2)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

### Dividing Complex Numbers

Dividing two complex numbers can be a little complicated. These problems are normally written in fraction form. If the denominator is purely imaginary, we can simply multiply the fraction by and simplify.

#### Examples

Perform the division. Write the quotient in the form a + bi , where a and b are real numbers.

•

•

When the divisor (denominator) is in the form a + bi , multiplying the fraction by will not work.

What does work is to multiply the fraction by the denominator’s conjugate over itself. This works because the product of any complex number and its conjugate is a real number. We will use the FOIL method in the numerator (if necessary) and the fact that (a + bi)(a − bi) = a 2 + b 2 in the denominator.

#### Examples

Write the quotient in the form a + bi , where a and b are real numbers.

### Complex Numbers on the Plane

There are reasons to write complex numbers in the form a + bi . One is that complex numbers are plotted in the plane (real numbers are plotted on the number line), where the x -axis becomes the real axis and the y -axis becomes the imaginary axis. The number 3 − 4 i is plotted in Figure 7.15.

Fig. 7.15

## Complex Numbers Practice Problems

#### Practice

For Problems 1–3, write the complex number in the form a + bi , where a and b are real numbers.

For Problems 4–15, perform the arithmetic. Write answers in the form a + bi , where a and b are real numbers.

1. 18 − 4 i + (−15) + 2 i
2. 5 + i + 5 − i
3. 7 + i + 12 + i

1. (15 + 3 i )(−2 + i

2. (3 + 2 i )(3 − 2 i

3. (8 − 10 i )(8 + 10 i

4. (1 − 9 i )(1 + 90)

For Problems 16–18, identify the complex conjugate.

5. 15 + 7 i

6. −3 + i

7. −9 i

For Problems 19–21, write the quotient in the form a + bi , where a and b are real numbers.

#### Solution

1. 18 − 4 i + (−15) + 2 i = 3 − 2 i

2. 5 + i + 5 − i = 10 + 0 i = 10
3. 7 + i + 12 + i = 19 + 2 i

9.  (2 i )(10 i ) = 20 i 2 = 20(−1) = −20

1. (15 + 3 i )(−2 + i ) = −30 + 15 i − 6 i + 3 i 2 = −30 + 9 i + 3(−1) = −33 + 9 i

2. (3 + 2 i )(3 − 2 i ) = 9 − 6 i + 6 i −4 i 2 = 9 − 4(−1) = 13(or 3 2 + 2 2 = 13)

3. (8 − 10 i )(8 + 10 i ) = 64 + 80 i − 80 i − 100 i 2 = 64 − 100(−1) = 164 (or 8 2 + 10 2 = 164)

4. (1 − 9 i )(1 + 9 i ) = 1 + 9 i − 9 i −81 i 2 = 1 − 81(−1) = 82(or 1 2 + 9 2 = 82)

5. The complex conjugate of 15 + 7 i is 15 − 7 i .

6. The complex conjugate of − 3 + i is − 3 − i .

7. The complex conjugate of −9 i is 9 i .

Practice problems for this concept can be found at: Polynomial Functions Practice Test.

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