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# 6.6: The Product & Quotient Theorems

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Determine the quotient theorem of complex numbers in polar form.
• Determine the product theorem of complex numbers in polar form.
• Solve everyday problems that require you to use the product and/or quotient theorem of complex numbers in polar form to obtain the correct solution.

## The Product Theorem

Multiplication of complex numbers in polar form is similar to the multiplication of complex numbers in standard form. However, to determine a general rule for multiplication, the trigonometric functions will be simplified by applying the sum/difference identities for cosine and sine. To obtain a general rule for the multiplication of complex numbers in polar from, let the first number be $r_1(\cos \theta_1 + i \sin \theta_1)$ and the second number be $r_2(\cos \theta_2 + i \sin \theta_2)$. The product can then be simplified by use of three facts: the definition $i^2 = -1$, the sum identity $\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)$, and the sum identity $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)$.

Now that the numbers have been designated, proceed with the multiplication of these binomials.

$& r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2)\\& r_1r_2(\cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2)\\& r_1r_2[(\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2)]\\& r_1r_2[\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)]$

Therefore:

$r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2)=r_1r_2[\cos (\theta_1+\theta_2)+i \sin(\theta_1+\theta_2)]$

## Quotient Theorem

Division of complex numbers in polar form is similar to the division of complex numbers in standard form. However, to determine a general rule for division, the denominator must be rationalized by multiplying the fraction by the complex conjugate of the denominator. In addition, the trigonometric functions must be simplified by applying the sum/difference identities for cosine and sine as well as one of the Pythagorean identities. To obtain a general rule for the division of complex numbers in polar from, let the first number be $r_1(\cos \theta_1 + i \sin \theta_1)$ and the second number be $r_2(\cos \theta_2 + i \sin \theta_2)$. The product can then be simplified by use of five facts: the definition $i^2 = -1$, the difference identity $\cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta)$, the difference identity $\sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin (\alpha - \beta)$, the Pythagorean identity, and the fact that the conjugate of $\cos \theta_2 + i \sin \theta_2$ is $\cos \theta_2 - i \sin \theta_2$.

$& \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}\\& \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)} \cdot \frac{(\cos \theta_2-i \sin \theta_2)}{(\cos \theta_2-i \sin \theta_2)}\\& \frac{r_1}{r_2} \cdot \frac{\cos \theta_1 \cos \theta_2-i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2-i^2 \sin \theta_1 \sin \theta_2}{\cos^2 \theta_2-i^2 \sin^2 \theta_2}\\& \frac{r_1}{r_2} \cdot \frac{(\cos \theta_1 \cos \theta_2+ \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2-\cos \theta_1 \sin \theta_2)}{\cos^2 \theta_2 + \sin^2 \theta_2}\\& \frac{r_1}{r_2}[\cos (\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]$

In general:

$\frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]$

## Using the Product and Quotient Theorems

The following examples illustrate the use of the product and quotient theorems.

Example 1: Find the product of the complex numbers $3.61(\cos 56.3^\circ + i \sin 56.3^\circ)$ and $1.41(\cos 315^\circ + i \sin 315^\circ)$

Solution: Use the Product Theorem, $r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2) = r_1r_2[\cos (\theta_1 + \theta_2) + i \sin (\theta_1 + \theta_2)]$.

$& \quad 3.61(\cos 56.3^\circ + i \sin 56.3^\circ) \cdot 1.41(\cos 315^\circ + i \sin 315^\circ)\\ & = (3.61)(1.41)[\cos(56.3^\circ + 315^\circ) + i \sin(56.3^\circ + 315^\circ)\\ & = 5.09(\cos 371.3^\circ + i \sin 371.3^\circ)\\ & = 5.09(\cos 11.3^\circ + i \sin 11.3^\circ)$

$^*$Note: Angles are expressed $0^\circ \le \theta \le 360^\circ$ unless otherwise stated.

Example 2: Find the product of $5 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4} \right ) \cdot \sqrt{3} \left (\cos \frac{\pi}{2}+i \sin \frac{\pi}{2} \right )$

Solution: First, calculate $r_1r_2=5 \cdot \sqrt{3}=5\sqrt{3}$ and $\theta =\theta_1+\theta_2=\frac{3\pi}{4}+\frac{\pi}{2}=\frac{5\pi}{4}$

$5\sqrt{3} \left (\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4} \right )$

Example 3: Find the quotient of $(\sqrt{3}-i) \div (2- i2\sqrt{3})$

Solution: Express each number in polar form.

$& \sqrt{3}-i && 2-i2\sqrt{3}\\& r_1=\sqrt{x^2+y^2} && r_2=\sqrt{x^2+y^2}\\& r_1=\sqrt{(\sqrt{3})^2+(-1)^2} && r_2 = \sqrt{(2)^2+(-2\sqrt{3})^2}\\& r_1=\sqrt{4}=2 && r_2=\sqrt{16}=4$

$& \frac{r_1}{r_2}=.5\\& \theta_1=\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) && \theta_2=\tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) && \theta=\theta_1-\theta_2\\& \theta_1=5.75959 \ rad. && \theta_2=5.23599 \ rad. && \theta=5.75959-5.23599\\&&&&& \theta=0.5236$

Now, plug in what we found to the Quotient Theorem.

$\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]=.5(\cos 0.5236+i \sin 0.5236)$

Example 4: Find the quotient of the two complex numbers $28 \angle 35^\circ$ and $14 \angle 24^\circ$

Solution:

$& \text{For} \ 28 \ \angle 35^\circ && \text{For} \ 14 \ \angle 24^\circ && \frac{r_1}{r_2}=\frac{28}{14}=2\\& r_1=28 && r_2=14 && \theta=\theta_1-\theta_2\\& \theta_1=35^\circ && \theta_2=24^\circ && \theta=35^\circ-24^\circ=11^\circ$

$\frac{r_1 \angle \theta_1}{r_2 \angle \theta_2} &= \frac{r_1}{r_2} \angle (\theta_1-\theta_2)\\&=2 \angle 11^\circ$

## Points to Consider

• We have performed the basic operations of arithmetic on complex numbers, but we have not dealt with any exponents or any roots of complex numbers. How might you calculate $(x + yi)^2$ or $\sqrt{r\angle \theta}$?
• How might you calculate the $n^{th}$ power or root of a complex number?

## Review Questions

1. Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.
1. $2 \angle 56^\circ, 7 \angle 113^\circ$
2. $3(\cos \pi + i \sin \pi), 10 \left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)$
3. $2+3i, -5+11i$
4. $6-i, -20i$
2. Part c from #1 was not in polar form. Mulitply the two complex numbers together without changing them into polar form. Which method do you think is easier?
3. Use the Product Theorem to find $4 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^2$.
4. The electric power (in watts) supplied to an element in a circuit is the product of the voltage $e$ and the current $i$ (in amps). Find the expression for the power supplied if $e=6.80 \angle 56.3^\circ$ volts and $i=7.05 \angle -15.8^\circ$ amps. Note: Use the formula $P = ei$.
5. Divide the following complex numbers. If they are not in polar form, change them before dividing. In
1. $\frac{2 \angle 56^\circ}{7 \angle 113^\circ}$
2. $\frac{10 \left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}$
3. $\frac{2+3i}{-5+11i}$
4. $\frac{6-i}{1-20i}$
6. Part c from #5 was not in polar form. Divide the two complex numbers without changing them into polar form. Which method do you think is easier?
7. Use the Product Theorem to find $4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^3$. Hint: use #3 to help you.
8. Using the Quotient Theorem determine $\frac{1}{4cis \frac{\pi}{6}}$.

1. $2 \angle 56^\circ, 7 \angle 113^\circ=(2)(7) \angle (56^\circ+113^\circ)=14 \angle 169^\circ$
2. $3(\cos \pi+i \sin \pi), 10 \left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)=(3)(10)cis \left(\pi+\frac{5\pi}{3}\right)=30cis \frac{8\pi}{3}=30cis \frac{2\pi}{3}$
3. $2+3i, -5+11i \rightarrow \ \text{change to polar}$$& x=2, y=3 && x=-5, y=11\\& r=\sqrt{2^2+3^2}=\sqrt{13} \approx 3.61 && r=\sqrt{(-5)^2+11^2}=\sqrt{146} \approx 12.08\\& \tan \theta =\frac{3}{2} \rightarrow \theta=56.31^\circ && \tan \theta=-\frac{11}{5} \rightarrow \theta=114.44^\circ$$(3.61)(12.08) \angle (56.31^\circ+114.44^\circ)=43.61 \angle 170.75^\circ$
4. $6-i, -20i \rightarrow \ \text{change to polar}$

$& x=6, y=-1 && x=0, y=-20\\& r=\sqrt{6^2+(-1)^2}=\sqrt{37} \approx 6.08 && r=\sqrt{0^2+(-20)^2}=\sqrt{40}=20\\& \tan \theta=-\frac{1}{6} \rightarrow \theta=350.54^\circ && \tan \theta=\frac{-20}{0}=und \rightarrow \theta=270^\circ$ $(6.08)(20) \angle (350.54^\circ+270^\circ)=121.6 \angle 620.54^\circ=121.6 \angle 260.54^\circ$

1. Without changing complex numbers to polar form, you mulitply by FOIL-ing. $(2+3i)(-5+11i)=-10+22i-15i+33i^2=-10-33+7i=-43+7i$ The answer is student opinion, but they seem about equal in the degree of difficulty.
2. $4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^2 &= 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \cdot 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\\&= 16\left(\cos \left(\frac{\pi}{4}+\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}+\frac{\pi}{4}\right)\right)\\&= 16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right)$
3. $P=(6.80)(7.05) \angle (56.3^\circ-15.8^\circ), P=47.9 \angle 40.5^\circ watts$
1. $\frac{2 \angle 56^\circ}{7 \angle 113^\circ}=\frac{2}{7} \angle (56^\circ-113^\circ)=\frac{2}{7} \angle -57^\circ$
2. $\frac{10\left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}=2 \left(\cos \left(\frac{5\pi}{3}-\pi\right)+i \sin \left(\frac{5\pi}{3}-\pi\right)\right)=2\left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)$
3. $\frac{2+3i}{-5+11i} \rightarrow \ \text{change each to polar}.$$& x=2, y=3 && x=-5, y=11\\& r=\sqrt{2^2+3^2}=\sqrt{13} \approx 3.61 && r=\sqrt{(-5)^2+11^2}=\sqrt{146} \approx 12.08\\& \tan \theta = \frac{3}{2} \rightarrow \theta = 56.31^\circ && \tan \theta=-\frac{11}{5} \rightarrow \theta=114.44^\circ$$\frac{3.61}{12.08} \angle (56.31^\circ-114.44^\circ)=0.30 \angle - 58.13^\circ$
4. $\frac{6-i}{1-20i} \rightarrow \ \text{change both to polar}$$& x=6, y=-1 && x=1, y=-20\\& r=\sqrt{6^2+(-1)^2}=\sqrt{37} \approx 6.08 && r=\sqrt{1^2+(-20)^2}=\sqrt{401}=20.02\\& \tan \theta=-\frac{1}{6} \rightarrow \theta=350.54^\circ && \tan \theta =\frac{-20}{1} \rightarrow \theta=272.68^\circ$$\frac{6.08}{20.02} \angle (350.54^\circ-272.86^\circ)=0.304 \angle 77.68^\circ$
4. $\frac{2+3i}{-5+11i}=\frac{2+3i}{-5+11i} \cdot \frac{-5-11i}{-5-11i}=\frac{-10-22i-15i+33}{25+121}=\frac{23-37i}{146}$ Again, this is opinion, but in general, using the polar form is “easier.”
5. $[4\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)]^3=[4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)]^2 \cdot 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right).$ From #3, $[4\left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)]^2=16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right)$. So, $[4\left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)]^3 &= 16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right) \cdot 4 \left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)\\&= 64\left(\cos \frac{3\pi}{4}i \sin \frac{3\pi}{4}\right)$
6. Even though 1 is not a complex number, we can still change it to polar form. $1 \rightarrow x=1, y=0$ $r &= \sqrt{1^2+0^2}=1 \qquad \\\tan \theta &= \frac{0}{1}=0 \rightarrow \theta = 0^\circ$ $\text{So}, \frac{1}{4cis\frac{\pi}{6}}=\frac{1cis0}{4cis\frac{\pi}{6}}=\frac{1}{4}cis \left(0-\frac{\pi}{6}\right)=\frac{1}{4}cis\left(-\frac{\pi}{6}\right).$

Feb 23, 2012

Apr 23, 2015