5.3.5: Product and Quotient Theorems

Simplified way to multiply and divide complex numbers.

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:

\(V=V_0e^{j\omega t}=V_0(\cos \omega t+j\sin \omega t)\)

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

\(r_2(\cos \theta_2+i\sin \theta_2)\)

Product and Quotient Theorems

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 \(z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)\), \(z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)\)

To multiply the two complex numbers in polar form:

\(\begin{aligned} z_1 \cdot z_2 &=r_1(\cos \theta_1+i \sin \theta_1)\cdot r_2(\cos \theta_2+i \sin \theta_2)\\ &=r_1r_2(\cos \theta_1+i \sin \theta_1)(\cos \theta_2+i \sin \theta_2) \\ &=r_1r_2\cdot (\cos \theta_1 \cos \theta_2+i \cos \theta_1 \sin \theta_2+i \sin \theta_1 \cos \theta_2+i^2 \sin \theta_1 \sin \theta_2) \\ &=r_1r_2(\cos \theta_1 \cos \theta_2 +i \cos \theta_1 \sin \theta_2+i \sin \theta_1 \cos \theta_2 −\sin \theta_1 \sin \theta_2)\\ &=r_1r_2 (\cos \theta_1 \cos \theta_2−\sin \theta_1 \sin \theta_2+i \cos \theta_1 \sin \theta_2+i \sin \theta_1 \cos \theta_2) \\ &=r_1r_2 ([\cos \theta_1 \cos \theta_2−\sin \theta_1 \sin \theta_2]+i[\cos \theta_1 \sin \theta_2+\sin \theta_1 \cos \theta_2]) \end{aligned}\)

(Use \(i^2 = -1\), gather like terms, factor out i, substitute the angle sum formulas for both sine and cosine)

\(z_1 \cdot z_2 =r_1r_2 cis \; [(\theta_1+\theta_2)]\)

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.

The 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 \(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\), then \(\dfrac{z_1}{z_2} =\dfrac{r_1}{r_2} \times \; cis \; [\theta_1−\theta_2]\)

Example \(\PageIndex{1}\)

Earlier, you were asked to identify the variable j in the following formula:

\(V=V_0e^{j\omega t}=V_0(\cos \omega t+j\sin \omega t)\)

Solution

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 \(r(\cos \theta+i\sin \theta)\).

Example \(\PageIndex{2}\)

Multiply \(z_1 \cdot z_2 \) where \(z_1 =2+2i\) and \(z_2 =1−\sqrt{3i}\).

Solution

For \(z_1 \),

\(\begin{aligned} r_1&=\sqrt{2^2+2^2} \\ &=\sqrt{8} \\ &=2\sqrt{2} \end{aligned}\)

and

\(\begin{aligned} \tan \theta_1&=\dfrac{2}{2} \\ \tan \theta_1&=1 \\ \theta_1&=\dfrac{\pi }{4}\end{aligned}\)

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

For \(z_2 \),

\(\begin{aligned} r_2&=\sqrt{1^2+(−\sqrt{3})^2} \\&=\sqrt{1+3} \\ &=\sqrt{4}\\ &=2\end{aligned}\)

and

\(\begin{aligned} \tan \theta_2 &=\dfrac{- \sqrt{3}}{1} \\ \theta_2&=\dfrac{5 \pi}{3} \end{aligned}\)

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

Substituting gives:

\(\begin{aligned} z_1 \cdot z_2 =2\sqrt{2} \times 2 \; cis \; \left[\dfrac{\pi}{4}+\dfrac{5\pi}{3} \right] \\ =4\sqrt{2} \; cis \; \left[\dfrac{23\pi }{12}\right] \end{aligned}\)

So we have

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

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

\(\begin{aligned} z_1 \times z_2 &=(2+2i)(1−\sqrt{3} i) \text{ or }\\ &=2−2\sqrt{3} i+2i−2\sqrt{3} i^2 \end{aligned}\)

Gathering like terms and using \(i^2 = -1\)

\(=(2+2\sqrt{3} )−(2\sqrt{3} +2)i\)

or

\(5.46−1.46i\)

Example \(\PageIndex{3}\)

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 \(\theta_1\), first find \(\tan \theta_{ref}=\left | \dfrac{y}{x} \right|\)

\(\begin{aligned} \tan \theta_{ref}&=\dfrac{(2\sqrt{3} )}{6} \\ \tan \theta_{ref}&=\dfrac{\sqrt{3} }{3} \\ \theta_{ref}&=\dfrac{\pi}{6} \end{aligned}\)

Since \(x > 0\) and \(y < 0\) we know that \(\theta_1\) is in the in the 4^th quadrant:

\(\theta_1=11\dfrac{\pi}{6}\)

For \(\theta_2\),

\(\begin{aligned} \tan \theta_{ref}&=\dfrac{(4\sqrt{3} )}{4} \\ \tan \theta_{ref}&=\sqrt{3} \\ \theta_{ref}&=\dfrac{\pi}{3} \end{aligned}\)

Since \(\theta_2\) is in the first quadrant,

\(\theta_2=\dfrac{\pi}{3}\)

Using polar multiplication,

\(\begin{aligned} z_1 \times z_2 &=4\sqrt{3} \times 8\left(\; cis \; \left[\dfrac{11\pi}{6}+\dfrac{\pi}{3}\right] \right) \\ z_1 \times z_2 &=32\sqrt{3} \left(\; cis \; \left[\dfrac{13 \pi}{6}\right] \right) \end{aligned}\)

subtracting \(2\pi\) from the augment:

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

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

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

Check:

\(\begin{aligned} (6−2\sqrt{3} i)(4+4\sqrt{3} i)&=24+24\sqrt{3} i−8\sqrt{3} i−24i^2\\ &=24+16\sqrt{3} i+24 \\ &=48+27.713i \end{aligned}\)

Example \(\PageIndex{4}\)

Using polar division, find the quotient of \(z_1 z_2 \) given that \(z_1 =5−5i\) and \(z_2 =−2\sqrt{3} −2i\).

Solution

For \(z_1 \): \(r_1=\sqrt{5^2+(−5)^2}\) or \( 5\sqrt{2} \) and \(\tan \theta_1=\dfrac{−5}{5}\), so \(\theta_1=\dfrac{7\pi}{4}\) (4th quadrant)

For \(z_2 \): \(r_2=\sqrt{(−2\sqrt{3} )^2+(−2)^2}\) or \(\sqrt{16}=4\) and \(\tan \theta_2=\dfrac{−2}{(−2\sqrt{3} )}\), so \(\theta_2=\dfrac{7 \pi}{6}\) (3^rd quadrant)

Using the formula, \(z_1 z_2 =r_1r_2\times \; cis \; [\theta_1−\theta_2]\) or

\(\begin{aligned} &=5\sqrt{2} 4\times \; cis \; \left[\dfrac{7\pi}{4}−\dfrac{7 \pi}{6}\right] \\ &=5\sqrt{2} 4\times \; cis \; \left[\dfrac{7 \pi}{12}\right] \\ &=5\sqrt{2} 4\left[\cos \dfrac{7 \pi}{12}+i \sin \dfrac{7 \pi}{12}\right]\\ &=1.768[−0.259+(0.966)i] \\ &=−0.458+1.708i \end{aligned}\)

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

\(\begin{aligned}
\dfrac{5-5 i}{-2 \sqrt{3}-2 i} \cdot \dfrac{-2 \sqrt{3}+2 i}{-2 \sqrt{3}+2 i}&=\dfrac{-10 \sqrt{3}+10 i+10 \sqrt{3} i-10 i^{2}}{(-2 \sqrt{3})^{2}-(2 i)^{2}} \\
&=\dfrac{-10 \sqrt{3}+10 i+10 \sqrt{3} i+10}{12+4} \\
&=\dfrac{(-10 \sqrt{3}+10)+(10+10 \sqrt{3}) i}{16} \\
&=\dfrac{(-17.3+10)+(10+17.3) i}{16} \\
&=\dfrac{(-7.3)+(27.3) i}{16} \qquad \text { or } \; -0.456+1.706 i
\end{aligned}\)

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

Example \(\PageIndex{5}\)

Find the product: \(\left(7\left(\dfrac{\pi}{6}\right)\right)\cdot \left(5 \left (−\dfrac{\pi}{4}\right) \right)\).

Solution

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

\(\begin{aligned} r_{1} \cdot r_{2} &\rightarrow 7 \cdot 5=35 && \text{By substitution and multplication}\\ \theta_{1}+\theta_{2} & \rightarrow\left(\dfrac{\pi}{6}\right)+\left(\dfrac{-\pi}{4}\right) && \text{Substitute} \\
&\left(\dfrac{2 \pi}{12}\right)+\left(\dfrac{-3 \pi}{12}\right) && \text{Find common denominators} \\ &\left(\dfrac{-\pi}{12}\right) && \text{Simplify} \\
&\therefore 35 \; cis \; \left(\dfrac{-\pi}{12}\right) && \text{is the product } \end{aligned}\)

Example \(\PageIndex{6}\)

Find the quotient: \(\dfrac{1+2i}{2−i}\).

Solution

First, find the quotient by polar multiplication:

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

\(\begin{aligned} \tan \theta_1&=\dfrac{2}{1} \\ \tan \theta_1&=2 \\ \theta_{ref}&=1.107 \text{ radians} \end{aligned}\)

since the angle is in the 1^st quadrant

\(\theta_1 = 1.107 \text{ radians}\)

for \(\theta_2\),

\(\begin{aligned} \tan \theta_2&=\dfrac{−1}{2} \\ \tan \theta_{ref}&=\dfrac{1}{2} \\ \theta_{ref}&=0.464 \text{ radians }\end{aligned}\)

since\(\theta_2\) is in the 4^th quadrant, between 4.712 (or \(\dfrac{3 \pi }{2}\)) and 6.282 radians (or \(2\pi \))

\(\theta_2 = 5.820 radians\)

Finally, using the division formula,

\(\begin{aligned} \dfrac{z_1 }{z_2} &=\dfrac{\sqrt{5} }{\sqrt{5} }[\; cis \; (1.107−5.820)] \\ \dfrac{z_1 }{z_2} &=[\; cis \; (−4.713)] \\ \dfrac{z_1 }{z_2}&=[\cos (−4.713)+i \sin (−4.713)] \\ \dfrac{z_1 }{z_2}&=[\cos (1.570)+i \sin (1.570)] \end{aligned}\)

If we assume that \(\dfrac{\pi }{2}=1.570\), then

\(\begin{aligned} &\approx \dfrac{z_1 }{z_2}=[\cos \left(\dfrac{\pi }{2}\right)+i \sin \left(\dfrac{\pi }{2}\right)] \\ \dfrac{z_1 }{z_2}&=0+1i=i \end{aligned}\)

Review

1. Find the product using polar form: \((2+2i)(\sqrt{3}−i)\)
2. \(2 cis \; (40)\cdot 4 cis \; (20)\)
3. Multiply: \(2\left(\cos \dfrac{\pi }{8}+i \sin \dfrac{\pi }{8}\right)\cdot 2 \left(\cos \dfrac{\pi }{10}+i \sin \dfrac{\pi }{10}\right)\)
4. \(\dfrac{2cis \; (80)}{6cis \; (200)}\)
5. Divide: \(3cis \; (130^{\circ}) \div 4cis \; (270^{\circ})\)

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

6. \(z_1 \cdot z_2\)
7. \(\left(\dfrac{z_1 }{z_2} \right)\)
8. \(\left(\dfrac{z_2 }{z_1 }\right)\)

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

9. \(z_1 z_2\)
10. \(\left(\dfrac{z_1 }{z_2} \right)\)
11. \(\left(\dfrac{z_2 }{z_1 }\right)\)
12. \((z_1 )^2\)
13. \((z_2 )^3\)

Find the products.

14. Find the product using polar form:\((2+2i)(\sqrt{3} −i)\)
15. \(2(\cos 40^{\circ}+i\sin 40^{\circ})\cdot 4(\cos 20^{\circ}+i\sin 20^{\circ})\)
16. \(2\left(\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}\right)\cdot 2\left(\cos \dfrac{\pi }{10}+i\sin \dfrac{\pi }{10}\right)\)

Find the quotients.

17. \(2(\cos 80^{\circ}+i\sin 80^{\circ})\div 6(\cos 200^{\circ}+i\sin 200^{\circ})\)
18. \(3cis \; (130^{\circ})\div 4cis \; (270^{\circ})\)

Review (Answers)

To see the Review answers, open this PDF file and look for section 4.9.

Vocabulary

Term	Definition
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 \sin gle real 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	The rectangular form of a point or a curve is given in terms of x and y and is graphed on the Cartesian plane.

Additional Resources

Video: Complex Conjugates Theorem - Overview

Practice: Product and Quotient Theorems