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Product and Quotient Theorems

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Complex numbers are found in real world calculations involving: quantum mechanics, signal analysis, fluid dynamics, control theory, and many other fields.

In electrical engineering, complex numbers are used for calculations involving impedance (the resistance to electric flow in a circuit).

Electrical engineers are familiar with the formula:

 V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right )

by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable j ?

r_2 (cos \theta _2 + i sin \theta _2)

Watch This

Embedded Video:

- James Sousa: The Product and Quotient of Complex Numbers in Trigonometric Form

Guidance

The 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:

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)

= r_1 r_2(\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1)(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)

= 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)

= 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)

= 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)

= 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])

(Use i 2 = -1 Gather like terms, factor out i, Substitute the angle sum formulas for both sine and cosine)

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

This last equation states that 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 is more concise 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 \frac{z_1} {z_2} = \frac{r_1} {r_2} \times \mbox{cis}\ [\theta_1 - \theta_2]

Example A

Multiply z_1 \cdot z_2 where z_1 = 2 + 2i and z_2 = 1 - \sqrt{3}i

Solution

For z 1 ,

r_1 = \sqrt{2^2 + 2^2}

= \sqrt{8}

= 2\sqrt{2}

and

\mbox{tan}\ \theta_1 = \frac{2} {2}

\mbox{tan}\ \theta_1 = 1

\theta_1 = \frac{\pi} {4}

Note that \theta_1 is in the first quadrant since a , and b > 0.

For z_2 ,

r_2 = \sqrt{1^2 + (-\sqrt{3})^2}

= \sqrt{1 + 3}

= \sqrt{4}

= 2

and

\mbox{tan}\ \theta_2 = \frac{-\sqrt{3}} {1},

\theta_2 = \frac{5\pi} {3}

Now we can use the formula z_1 \cdot z_2 = r_1 \cdot r_2 cis \left( \theta_1 +\theta_2 \right)

Substituting gives:

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

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

So we have

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

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

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

= 2 - 2\sqrt{3}i + 2i - 2\sqrt{3}i^2

Gathering like terms and using i 2 = -1

= (2 + 2\sqrt{3}) - (2\sqrt{3} + 2)i

or

5.46 - 1.46i

Example B

Using polar multiplication, find the product (6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i)

Solution

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

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

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

For θ 1 , first find \mbox{tan}\ \theta_{ref} = \left |\frac{y} {x}\right |

\mbox{tan}\ \theta_{ref} = \frac{(2\sqrt{3})} {6}

\mbox{tan}\ \theta_{ref} = \frac{\sqrt{3}} {3}

\theta_{ref} = \frac{\pi} {6}.

Since x > 0 and y < 0 we know that θ 1 is in the in the 4 th quadrant:

\theta_1 = \frac{11\pi} {6}

For θ 2 ,

\mbox{tan}\ \theta_{ref} = \frac{(4\sqrt{3})} {4}

\mbox{tan}\ \theta_{ref} = \sqrt{3},

\theta_{ref} = \frac{\pi} {3}

Since θ 2 is in the first quadrant,

\theta_2 = \frac{\pi} {3}

Using polar multiplication,

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

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

subtracting 2π from the augment:

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

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

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

Check:

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

= 24 + 16\sqrt{3}i + 24

= 48 + 27.713i

Example C

Using polar division, find the quotient of \frac{z_1} {z_2}

Solution

Given that

z_1 = 5 - 5i

z_2 = -2\sqrt{3} - 2i

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

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

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

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

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

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

= 1.768[-0.259 + (0.966)i]

= -0.458 + 1.708i

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

\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}

= \frac{-10\sqrt{3} + 10i + 10\sqrt{3}i+ 10} {12 + 4}

= \frac{(-10\sqrt{3} + 10) + (10 + 10\sqrt{3})i} {16}

= \frac{(-17.3 + 10) + (10 + 17.3)i} {16}

= \frac{(-7.3) + (27.3)i} {16}\ \mbox{or}

-0.456 + 1.706i

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

Concept question wrap-up

Electrical engineers are familiar with the formula:

 V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right )

by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable j ?

In electrical calculations, the letter I is commonly used to denote current , therefore imaginary numbers are identified with a j .

Note the similar usage of i in r (cos \theta + i sin \theta)

Vocabulary

The product theorem is used for the multiplication of complex numbers.

The quotient theorem is used for the division of complex numbers.

Complex conjugates are pairs of complex binomials which result in a single real number product.

Guided Practice

1) Find the product: \left( 7 \left( \frac{\pi}{6} \right) \right) \bullet \left( 5 \left( \frac{-\pi}{4} \right) \right)

2) Find the quotient: \frac{1 + 2i}{2 - i}

Solutions

1) This one is easier than it looks: Recall z_1 \cdot z_2 = r_1 \cdot r_2 cis (\theta_1 + \theta_2)

a) r_1 \cdot r_2 \to 7 \cdot 5 = 35 ....... By substitution and multplication
b) \theta_1 + \theta_2 \to \left( \frac{\pi}{6} \right) + \left( \frac{-\pi}{4} \right) ..... Substitute
\left( \frac{2\pi}{12} \right) + \left( \frac{-3\pi}{12} \right) ..... Find common denominators
\left( \frac{-\pi}{12} \right) ..... Simplify

\therefore 35 cis \left( \frac{-\pi}{12} \right) is the product

2) First finding the quotient by polar multiplication:

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

\mbox{tan}\ \theta_1 = \frac{2} {1}

\mbox{tan}\ \theta_1 = 2

\theta_{ref} = 1.107\ \mbox{radians}

since the angle is in the 1 st quadrant

θ 1 = 1.107 radians

for \theta_2 ,

\mbox{tan}\ \theta_2 = \frac{-1} {2}

\mbox{tan}\ \theta_{ref} = \frac{1} {2}

\theta_{ref} = 0.464\ \mbox{radians}

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

θ 2 = 5.820 radians

Finally, using the division formula,

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

\frac{z_1} {z_2} = [\mbox{cis}\ (-4.713)]

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

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

If we assume that \frac{\pi} {2} = 1.570 , then

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

\frac{z_1} {z_2} = 0 + 1i = i

Practice

  1. Find the product using polar form: (2 + 2i)(\sqrt{3} - i)
  2. 2  cis (40) \bullet 4 cis (20)
  3. Multiply: 2 \left (\mbox{cos}\ \frac{\pi} {8} + i\ \mbox{sin}\ \frac{\pi} {8}\right ) \bullet 2 \left (\mbox{cos}\ \frac{\pi} {10} + i\ \mbox{sin}\ \frac{\pi} {10}\right )
  4. \frac{2 cis (80)}{6 cis (200)}
  5. Divide: 3cis(130 o ) ÷ 4cis(270 o )

If  z_1 = 7 \left( \frac{\pi}{6} \right) and z_2 = 5 \left( \frac{-\pi}{4} \right) find:

  1. z_1 \cdot z_2
  2. \left( \frac{z_1}{z_2} \right)
  3. \left( \frac{z_2}{z_1} \right)

If z_1 = 8 \left( \frac{\pi}{3} \right) and  z_2 = 5 \left(\frac{\pi}{6} \right) find:

  1. z_1 z_2
  2. \left( \frac{z_1}{z_2} \right)
  3. \left( \frac{z_2}{z_1} \right)
  4. (z_1)^2
  5. (z_2)^3

Find the products

  1. Find the product using polar form: (2 + 2i)(\sqrt3 - i)
  2. 2 (cos 40^o + i sin 40^o) \bullet 4(cos 20^o + i sin 20^o)
  3.  2 \left(cos \frac{\pi}{8} + i sin \frac{\pi}{8} \right) \bullet 2 \left(cos \frac{\pi}{10} + i sin \frac{\pi}{10} \right)

Find the quotients

  1.  2(cos 80^o + i sin 80^o) \div 6(cos 200^o + i sin 200^o)
  2.  3cis(130^o) \div 4cis(270^o)

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