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Product and Quotient Theorems

Multiplying and dividing complex numbers in polar form.

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Practice Product and Quotient Theorems
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Product and Quotient Theorems

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:

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

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

\begin{align*}r_2 (cos \theta _2 + i sin \theta _2)\end{align*}

Watch This

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


The Product Theorem

Since complex numbers can be transformed to 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 \begin{align*}i\end{align*}, 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*}

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 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*}

Example A

Multiply \begin{align*}z_1 \cdot z_2\end{align*} where \begin{align*}z_1 = 2 + 2i\end{align*} 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*}


\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 \begin{align*}\theta_1\end{align*} is in the first quadrant since a, and b > 0.

For \begin{align*}z_2\end{align*},

\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*}


\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 \begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2 cis \left( \theta_1 +\theta_2 \right)\end{align*}

Substituting gives:

\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*}


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

Example B

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


\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 C

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


For \begin{align*}z_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}i+ 10} {12 + 4}\end{align*}

\begin{align*}= \frac{(-10\sqrt{3} + 10) + (10 + 10\sqrt{3})i} {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.

Concept question wrap-up:

Electrical engineers are familiar with the formula:

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

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 \begin{align*}r (cos \theta + i sin \theta)\end{align*}


Guided Practice

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

2) Find the quotient: \begin{align*}\frac{1 + 2i}{2 - i}\end{align*}


1) This one is easier than it looks: Recall \begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2\end{align*} \begin{align*}cis (\theta_1 + \theta_2)\end{align*}

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

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

2) 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 = i\end{align*}

Explore More

  1. Find the product using polar form: \begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}
  2. \begin{align*}2\end{align*} \begin{align*} cis (40) \bullet\end{align*} \begin{align*}4 \end{align*} \begin{align*}cis (20)\end{align*}
  3. Multiply: \begin{align*}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 )\end{align*}
  4. \begin{align*}\frac{2 cis (80)}{6 cis (200)}\end{align*}
  5. Divide: 3cis(130o) ÷ 4cis(270o)

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

  1. \begin{align*}z_1 \cdot z_2\end{align*}
  2. \begin{align*}\left( \frac{z_1}{z_2} \right)\end{align*}
  3. \begin{align*}\left( \frac{z_2}{z_1} \right)\end{align*}

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

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

Find the products.

  1. Find the product using polar form:\begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}
  2. \begin{align*}2 (cos 40^o + i sin 40^o) \bullet 4(cos 20^o + i sin 20^o)\end{align*}
  3. \begin{align*} 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)\end{align*}

Find the quotients.

  1. \begin{align*} 2(cos 80^o + i sin 80^o) \div 6(cos 200^o + i sin 200^o)\end{align*}
  2. \begin{align*} 3cis(130^o) \div 4cis(270^o)\end{align*}


Complex Conjugate

Complex Conjugate

Complex conjugates are pairs of complex binomials. The complex conjugate of a+bi is a-bi. When complex conjugates are multiplied, the result is a single real number.
complex number

complex number

A complex number is the sum of a real number and an imaginary number, written in the form a + bi.
rectangular form

rectangular form

The rectangular form of a point or a curve is given in terms of x and y and is graphed on the Cartesian plane.

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