# 4.7: Product and Quotient Theorems

**At Grade**Created by: CK-12

## Learning Objectives

- Use the rcis form of complex numbers to multiply two complex numbers
- Use the rcis form of complex numbers to divide two complex numbers
- Use the rectangular form and technological tools to check the results of multiplication and division with complex numbers

## Product Theorem

Since complex numbers can be transformed in polar form, the multiplication of complex numbers can also be done in polar form. Suppose we know *z*_{1} = *r*_{1} (cos *θ*_{1} + *i* sin *θ*_{1}) *z*_{2} = *r*_{2} (cos *θ*_{2} + *i* sin *θ*_{2})

To multiply the two complex numbers in polar form:

(Use *i*^{2} = -1 Gather like terms Factor out i Substitute the angle sum formulas for both sine and cosine)

In this last equation the product of two complex numbers in polar form can be obtained by multiplying the polar r values of each of the complex numbers and then multiplying that value by cis of the sum of each of the two angles of the individual complex numbers. This appears to be simpler than the rectangular form for multiplication of complex numbers.

## Quotient Theorem

Dividing complex numbers in polar form can be shown using a similar proof that was used to show multiplication of complex numbers. Here we omit the proof and give the result. For *z*_{1} = *r*_{1}(cos *θ*_{1} + *i* sin *θ*_{1}) and *z*_{2} = *r*_{2}(cos *θ*_{2} + *i* sin *θ*_{2}), then

## Using the Product and Quotient Theorem

**Example:** Multiply *z*_{1}*z*_{2} where *z*_{1} = 2 + 2*i* and

For *z*_{1},

and

Note that θ1 is in the first quadrant since *a*, and *b* > 0.

For *z*_{2},

and

Now we can use the formula, *z*_{1} ∙ *z*_{2} = *r*_{1}*r*_{2} cis [(*θ*_{1} + *θ*_{1})]. Substituting

So we have

Re-writing in approximate decimal form:

5.656 (0.966 – 0.259i)

5.46 - 1.46i

If the problem was done using only rectangular units then

Gathering like terms and using i^{2} = -1

or

**Example:** Using polar multiplication, find the product

Let and

and

and

For *θ*_{1}, first find

Since *x* > 0 and *y* < 0 we know that *θ*_{1} is in the in the 4^{th} quadrant:

For *θ*_{2},

Since *θ*_{2} is in the first quadrant,

Using polar multiplication,

subtracting 2π from the augment:

or in expanded form:

In decimal form this becomes: 55.426(0.866 + 0.500i) or 48 + 27.713i

Check:

**Example:** Using polar division find the quotient of , given that

for or and , so (4th quadrant)

for or and , so (3^{rd} quadrant)

Using the formula, or

Check by using the complex conjugate to do the division in rectangular form:

The two radically different approaches yield the same answer. The small difference between the two answers is a result of decimal rounding error.

**Example:** Find the quotient:

First finding the quotient by polar multiplication:

since the angle is in the 1^{st} quadrant

θ_{1} = 1.107 radians

for ,

since *θ*_{2} is in the 4^{th} quadrant, between 4.712 and 6.282 radians (or 2*π*)

θ_{2} = 5.820 *radians*

Finally, using the division formula,

If we assume that , then

To check, multiply numerator and denominator by the denominator’s complex conjugate:

## Applications, Technological Tools

After utilizing product or quotient laws, answers can be checked using a graphing calculator. Convert the given values into *a* + *bi* form and then put the calculator into “pol” mode. Perform the operation.

## Lesson Summary

There are two distinct methods when multiplying or dividing two complex numbers- one utilizes the complex number written in a + bi form, while the other uses the polar form. However, there is really only one practical method for finding a power or a root of a complex number and that is utilizing the polar form, or cis form, for the complex number.

## Points to Consider

Some students may be tempted to ask why is this new method presented- it is longer and (in this example) messier when there is an easier, less messier way of doing it? To give encouragement to learn alternative methods for solving problems, let’s look at science history, in particular Einstein and the Theory of Special Relativity.

Prior to 1905 when Special Relativity was published, motion was understood by Newton’s Laws of Motion and Galileo’s kinematics equations. For over 300 years whenever scientists used these laws to explain phenomena that involved motion their results were just about perfect. Yet by the beginning of the 20^{th} Century, Einstein glimpsed at some discrepancies within the laws of motion and the laws of electro-magnetism and realized something might be wrong with the way physicists were looking at the world. He developed a method that would bring these powerful laws together without discrepancies. What came out of his efforts was the Theory of Special Relativity. A cynic might have asked at that time, what was so important about learning a newer, more complicated method, when the older and reliable methods worked just as well? The answer for understanding Relativity was, yes, the new method yielded the identical answers as the older method, BUT the newer method was able to answer modern questions, not asked of the older methods. In other words it could do all that the older methods can do but also more. This idea has several examples in science and mathematics and is known as the *Correspondence Principle.*

This principle is true of the method of multiplying complex numbers. As seen, using the newer method yields identical results as the older method. After becoming proficient in using the newer method, problems will be presented that when using this newer method, will take much less time then the older method took, that is if the older method could even do the problem!

## Review Questions

- a. Find the product using polar form: b. Multiply: 2(cos 40
^{o}+*i*sin 40^{o}) × 4(cos 20^{o}+*i*sin 20^{o}) c. Multiply: d. Divide: 2(cos 80^{o}+*i*sin 80^{o}) ÷ 6(cos 200^{o}+*i*sin 200^{o}) e. Divide: 3cis(130^{o}) ÷ 4cis(270^{o})

## Review Answers

- a. b. 8 cis (60
^{o}) c. d. e.