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# 4.7: Product and Quotient Theorems

Difficulty Level: 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 z1 = r1 (cos θ1 + i sin θ1) z2 = r2 (cos θ2 + i sin θ2)

To multiply the two complex numbers in polar form:

\begin{align*}z_1 \cdot z_2 = r_1 (\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1) \cdot r_2(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)\end{align*}

\begin{align*}= r_1 r_2(\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1)(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)\end{align*}

\begin{align*}= r_1 r_2 \cdot (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2 + i^2\ \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2)\end{align*}

\begin{align*}= r_1 r_2 (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2\ + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2\ - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2)\end{align*}

\begin{align*}= r_1 r_2\ (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2)\end{align*}

\begin{align*}= r_1 r_2\ ([\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2] + i[\mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2])\end{align*}

(Use i2 = -1 Gather like terms Factor out i Substitute the angle sum formulas for both sine and cosine)

\begin{align*}z_1 \cdot z_2 = r_1r_2\ \mbox{cis}\ [(\theta_1 + \theta_2)]\end{align*}

\begin{align*}z_1 \cdot z_2 = r_1r_2\ \mbox{cis}\ [(\theta_1 + \theta_2)]\end{align*}

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 z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2), then \begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \times \mbox{cis}\ [\theta_1 - \theta_2]\end{align*}

## Using the Product and Quotient Theorem

Example: Multiply z1\begin{align*}\cdot\end{align*}z2 where z1 = 2 + 2i and \begin{align*}z_2 = 1 - \sqrt{3}i\end{align*}

For z1,

\begin{align*}r_1 = \sqrt{2^2 + 2^2}\end{align*}

\begin{align*}= \sqrt{8}\end{align*}

\begin{align*}= 2\sqrt{2}\end{align*}

and

\begin{align*}\mbox{tan}\ \theta_1 = \frac{2} {2}\end{align*}

\begin{align*}\mbox{tan}\ \theta_1 = 1\end{align*}

\begin{align*}\theta_1 = \frac{\pi} {4}\end{align*}

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

For z2,

\begin{align*}r_2 = \sqrt{1^2 + (-\sqrt{3})^2}\end{align*}

\begin{align*}= \sqrt{1 + 3}\end{align*}

\begin{align*}= \sqrt{4}\end{align*}

\begin{align*}= 2\end{align*}

and

\begin{align*}\mbox{tan}\ \theta_2 = \frac{-\sqrt{3}} {1},\end{align*}

\begin{align*}\theta_2 = \frac{5\pi} {3}\end{align*}

Now we can use the formula, z1z2 = r1r2 cis [(θ1 + θ1)]. Substituting

\begin{align*}z_1 \cdot z_2 = 2\sqrt{2} \times 2\ \mbox{cis}\ \left [\frac{\pi} {4} + \frac{5\pi} {3}\right ]\end{align*}

\begin{align*}= 4\sqrt{2}\ \mbox{cis}\ \left [\frac{23\pi} {12}\right ]\end{align*}

So we have

\begin{align*}z_1 \cdot z_2 = 4\sqrt{2} \left (\mbox{cos}\ \frac{23\pi} {12} + i\ \mbox{sin}\ \frac{23 \pi} {12}\right )\end{align*}

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

\begin{align*}z_1 \times z_2 = (2 + 2i)(1 - \sqrt{3}i) \ \mbox{or}\end{align*}

\begin{align*}= 2 - 2\sqrt{3}i + 2i - 2\sqrt{3}i^2\end{align*}

Gathering like terms and using i2 = -1

\begin{align*}= (2 + 2\sqrt{3}) - (2\sqrt{3} + 2)i\end{align*}

or

\begin{align*}5.46 - 1.46i\end{align*}

Example: Using polar multiplication, find the product \begin{align*}(6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i)\end{align*}

Let \begin{align*}z_1 = 6 - 2\sqrt{3}i\end{align*} and \begin{align*}z_2 = 4 + 4\sqrt{3}i\end{align*}

\begin{align*}r_1 = \sqrt{(6)^2 - (2\sqrt{3})^2}\end{align*} and \begin{align*}r_2 = \sqrt{(4)^2 + (4\sqrt{3})^2}\end{align*}

\begin{align*}r_1 = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}\end{align*} and \begin{align*}r_2 = \sqrt{16 + 48} = \sqrt{64} = 8\end{align*}

For θ1, first find \begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{y} {x}\right |\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{(2\sqrt{3})} {6}\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{\sqrt{3}} {3}\end{align*}

\begin{align*}\theta_{ref} = \frac{\pi} {6}.\end{align*}

Since x > 0 and y < 0 we know that θ1 is in the in the 4th quadrant:

\begin{align*}\theta_1 = \frac{11\pi} {6}\end{align*}

For θ2,

\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{(4\sqrt{3})} {4}\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \sqrt{3},\end{align*}

\begin{align*}\theta_{ref} = \frac{\pi} {3}\end{align*}

Since θ2 is in the first quadrant,

\begin{align*}\theta_2 = \frac{\pi} {3}\end{align*}

Using polar multiplication,

\begin{align*}z_1 \times z_2 = 4\sqrt{3} \times 8\left (\mbox{cis}\ \left [\frac{11\pi} {6} + \frac{\pi} {3}\right ]\right )\end{align*}

\begin{align*}z_1 \times z_2 = 32\sqrt{3} \left (\mbox{cis}\ \left[\frac{13\pi} {6}\right ]\right )\end{align*}

subtracting 2π from the augment:

\begin{align*}z_1 \times z_2 = 32\sqrt{3} \left (\mbox{cis}\ \left [\frac{\pi} {6}\right ]\right )\end{align*}

or in expanded form: \begin{align*}32\sqrt{3} \left (\mbox{cos}\ \left [\frac{\pi} {6}\right ] + i\ \mbox{sin}\ \left [\frac{\pi} {6}\right ]\right )\end{align*}

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

Check:

\begin{align*}(6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i) = 24 + 24\sqrt{3}i - 8\sqrt{3}i - 24i^2\end{align*}

\begin{align*}= 24 + 16\sqrt{3}i + 24\end{align*}

\begin{align*}= 48 + 27.713i\end{align*}

Example: Using polar division find the quotient of \begin{align*}\frac{z_1} {z_2}\end{align*}, given that

\begin{align*}z_1 = 5 - 5i\end{align*}

\begin{align*}z_2 = -2\sqrt{3} - 2i\end{align*}

for \begin{align*}z_1 : r_1 = r_1 = \sqrt{5^2 + (-5)^2}\end{align*} or \begin{align*}5\sqrt{2}\end{align*} and \begin{align*}\mbox{tan}\ \theta_1 = \frac{-5} {5}\end{align*}, so \begin{align*}\theta_1 = \frac{7\pi} {4}\end{align*} (4th quadrant)

for \begin{align*}z_2 : r_2 = \sqrt{(-2\sqrt{3})^2 + (-2)^2}\end{align*} or \begin{align*}\sqrt{16} = 4\end{align*} and \begin{align*}\mbox{tan}\ \theta_2 = \frac{-2} {(-2\sqrt{3})}\end{align*}, so \begin{align*}\theta_2 = \frac{7\pi} {6}\end{align*} (3rd quadrant)

Using the formula, \begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \times \mbox{cis}\ [\theta_1 - \theta_2]\end{align*} or

\begin{align*}= \frac{5\sqrt{2}} {4} \times \mbox{cis}\ \left [\frac{7\pi} {4} - \frac{7\pi} {6}\right ]\end{align*}

\begin{align*}= \frac{5\sqrt{2}} {4} \times \mbox{cis}\ \left [\frac{7\pi} {12}\right ]\end{align*}

\begin{align*}= \frac{5\sqrt{2}} {4} \left [\mbox{cos}\ \frac{7\pi} {12} + i\ \mbox{sin}\ \frac{7\pi} {12}\right ]\end{align*}

\begin{align*}= 1.768[-0.259 + (0.966)i]\end{align*}

\begin{align*}= -0.458 + 1.708i\end{align*}

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

\begin{align*}\frac{5 - 5i} {-2\sqrt{3} - 2i} \cdot \frac{-2\sqrt{3} + 2i} {-2 \sqrt{3} + 2i} = \frac{-10\sqrt{3} + 10i + 10\sqrt{3}i - 10i^2} {(-2\sqrt{3})^2 - (2i)^2}\end{align*}

\begin{align*}= \frac{-10\sqrt{3} + 10i + 10\sqrt{3}+ 10} {12 + 4}\end{align*}

\begin{align*}= \frac{(-10\sqrt{3} + 10) + (10 + 10\sqrt{3})} {16}\end{align*}

\begin{align*}= \frac{(-17.3 + 10) + (10 + 17.3)i} {16}\end{align*}

\begin{align*}= \frac{(-7.3) + (27.3)i} {16}\ \mbox{or}\end{align*}

\begin{align*}-0.456 + 1.706i\end{align*}

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: \begin{align*}\frac{1 + 2i} {2 - i}\end{align*}

First finding the quotient by polar multiplication:

\begin{align*}r_1 = \sqrt{(1)^2 + (2)^2} = \sqrt{5} \qquad r_2 = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}\end{align*}

\begin{align*}\mbox{tan}\ \theta_1 = \frac{2} {1}\end{align*}

\begin{align*}\mbox{tan}\ \theta_1 = 2\end{align*}

\begin{align*}\theta_{ref} = 1.107\ \mbox{radians}\end{align*}

since the angle is in the 1st quadrant

θ1 = 1.107 radians

for \begin{align*}\theta_2\end{align*},

\begin{align*}\mbox{tan}\ \theta_2 = \frac{-1} {2}\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{1} {2}\end{align*}

\begin{align*}\theta_{ref} = 0.464\ \mbox{radians}\end{align*}

since θ2 is in the 4th quadrant, between 4.712 \begin{align*}\left (\mbox{or}\ \frac{3\pi} {2}\right )\end{align*} and 6.282 radians (or 2π)

θ2 = 5.820 radians

Finally, using the division formula,

\begin{align*}\frac{z_1} {z_2} = \frac{\sqrt{5}} {\sqrt{5}} [\mbox{cis}\ (1.107 - 5.820)]\end{align*}

\begin{align*}\frac{z_1} {z_2} = [\mbox{cis}\ (-4.713)]\end{align*}

\begin{align*}\frac{z_1} {z_2} = [\mbox{cos}\ (-4.713) + i\ \mbox{sin}\ (-4.713)]\end{align*}

\begin{align*}\frac{z_1} {z_2} = [\mbox{cos}\ (1.570) + i\ \mbox{sin}\ (1.570)]\end{align*}

If we assume that \begin{align*}\frac{\pi} {2} = 1.570\end{align*}, then

\begin{align*}\approx \frac{z_1} {z_2} = \left [\mbox{cos}\ \left (\frac{\pi} {2}\right ) + i\ \mbox{sin}\ \left (\frac{\pi} {2}\right )\right ]\end{align*}

\begin{align*}\frac{z_1} {z_2} = 0 + 1i = 1\end{align*}

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

\begin{align*}\frac{1 + 2i} {2 - i} \cdot \frac{2 + i} {2 + i} = \frac{2 + i + 4i + 2i^2} {4 + 2i - 2i - i^2}\end{align*}

\begin{align*}= \frac{2 + 5i - 2} {4 + 1}\end{align*}

\begin{align*}= \frac{5i} {5}\end{align*}

\begin{align*}= i\end{align*}

## 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 20th 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

1. a. Find the product using polar form: \begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*} b. Multiply: 2(cos 40o + i sin 40o) × 4(cos 20o + i sin 20o) c. Multiply: \begin{align*}2 \left (\mbox{cos}\ \frac{\pi} {8} + i\ \mbox{sin}\ \frac{\pi} {8}\right ) \times 2 \left (\mbox{cos}\ \frac{\pi} {10} + i\ \mbox{sin}\ \frac{\pi} {10}\right )\end{align*} d. Divide: 2(cos 80o + i sin 80o) ÷ 6(cos 200o + i sin 200o) e. Divide: 3cis(130o) ÷ 4cis(270o)

1. a. \begin{align*}4\sqrt{2} (\mbox{cos}\ 15^\circ + i\ \mbox{sin}\ 15^\circ)\end{align*} b. 8 cis (60o) c. \begin{align*}4\ \mbox{cis}\ \left (\frac{9 \pi} {40}\right )\end{align*} d. \begin{align*}\frac{1} {3} \mbox{cis}\ (240^\circ)\end{align*} e. \begin{align*}\frac{3} {4} \mbox{cis}\ (220^\circ)\end{align*}

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Date Created:
Feb 23, 2012