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# 6.10: Product Theorem

Difficulty Level: At Grade Created by: CK-12
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Practice Product Theorem

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What if you were given two complex numbers in polar form, such as 2(cosπ2+isinπ2),7(cos3π2+isin3π2)\begin{align*}2\left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right), 7 \left(\cos \frac{3\pi}{2}+i \sin \frac{3\pi}{2}\right)\end{align*} and asked to multiply them? Would you be able to do this? How long would it take you?

After completing this Concept, you'll know the Product Theorem, which will make it easier to multiply complex numbers.

### Watch This

In the first part of this video you'll learn about the product of complex numbers in trigonometric form.

### Guidance

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 r1(cosθ1+isinθ1)\begin{align*}r_1(\cos \theta_1 + i \sin \theta_1)\end{align*} and the second number be r2(cosθ2+isinθ2)\begin{align*}r_2(\cos \theta_2 + i \sin \theta_2)\end{align*}. The product can then be simplified by use of three facts: the definition i2=1\begin{align*}i^2 = -1\end{align*}, the sum identity cosαcosβsinαsinβ=cos(α+β)\begin{align*}\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)\end{align*}, and the sum identity sinαcosβ+cosαsinβ=sin(α+β)\begin{align*}\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)\end{align*}.

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

r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)r1r2(cosθ1cosθ2+icosθ1sinθ2+isinθ1cosθ2+i2sinθ1sinθ2)r1r2[(cosθ1cosθ2sinθ1sinθ2)+i(sinθ1cosθ2+cosθ1sinθ2)]r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}& 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)]\end{align*}

Therefore:

r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}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)]\end{align*}

We can use this general formula for the product of complex numbers to perform computations.

#### Example A

Find the product of the complex numbers 3.61(cos56.3+isin56.3)\begin{align*}3.61(\cos 56.3^\circ + i \sin 56.3^\circ)\end{align*} and 1.41(cos315+isin315)\begin{align*}1.41(\cos 315^\circ + i \sin 315^\circ)\end{align*}

Solution: Use the Product Theorem, r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}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)]\end{align*}.

3.61(cos56.3+isin56.3)1.41(cos315+isin315)=(3.61)(1.41)[cos(56.3+315)+isin(56.3+315)=5.09(cos371.3+isin371.3)=5.09(cos11.3+isin11.3)\begin{align*}& \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)\end{align*}

\begin{align*}^*\end{align*}Note: Angles are expressed 0θ360\begin{align*}0^\circ \le \theta \le 360^\circ\end{align*} unless otherwise stated.

#### Example B

Find the product of 5(cos3π4+isin3π4)3(cosπ2+isinπ2)\begin{align*}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 )\end{align*}

Solution: First, calculate r1r2=53=53\begin{align*}r_1r_2=5 \cdot \sqrt{3}=5\sqrt{3}\end{align*} and θ=θ1+θ2=3π4+π2=5π4\begin{align*}\theta =\theta_1+\theta_2=\frac{3\pi}{4}+\frac{\pi}{2}=\frac{5\pi}{4}\end{align*}

\begin{align*}5\sqrt{3} \left (\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4} \right )\end{align*}

#### Example C

Find the product of the numbers \begin{align*}r_1 = 1 + i\end{align*} and \begin{align*}r_2 = \sqrt{3} - i\end{align*} by first converting them to trigonometric form.

Solution:

First, convert \begin{align*}1 + i\end{align*} to polar form:

\begin{align*} r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}\\ \theta_1 = \arctan \frac{1}{1}= \frac{\pi}{4}\\ \end{align*}

And now do the same with \begin{align*}\sqrt{3} - i\end{align*}:

\begin{align*} r_2 = \sqrt{\sqrt{3}^2 + (-1)^2} = 2\\ \theta_2 = \arctan -\frac{1}{\sqrt{3}} = -\frac{\pi}{6}\\ \end{align*}

And now substituting these values into the product theorem:

\begin{align*}r_1 \times r_2 =\\ (2)(\sqrt{2}) \left( \cos \left( -\frac{\pi}{6} + \frac{\pi}{4} \right) + i\sin \left( -\frac{\pi}{6} + \frac{\pi}{4} \right) \right)\\ (2)(\sqrt{2}) \left( \cos \left( \frac{\pi}{12} \right) + i\sin \left( \frac{\pi}{12} \right) \right) \end{align*}

### Vocabulary

Product Theorem: The product theorem is a theorem showing a simplified way to multiply complex numbers.

### Guided Practice

1. Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.

\begin{align*}2 \angle 56^\circ, 7 \angle 113^\circ\end{align*}

2. Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.

\begin{align*}3(\cos \pi + i \sin \pi), 10 \left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)\end{align*}

3. Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.

\begin{align*}2+3i, -5+11i\end{align*}

Solutions:

1. \begin{align*}2 \angle 56^\circ, 7 \angle 113^\circ=(2)(7) \angle (56^\circ+113^\circ)=14 \angle 169^\circ\end{align*}

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

3. \begin{align*}2+3i, -5+11i \rightarrow \ \text{change to polar}\end{align*}\begin{align*}& 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\end{align*}\begin{align*}(3.61)(12.08) \angle (56.31^\circ+114.44^\circ)=43.61 \angle 170.75^\circ\end{align*}

### Concept Problem Solution

Since you want to multiply

\begin{align*}2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}), 7 \left(\cos \frac{3\pi}{2}+i \sin \frac{3\pi}{2}\right)\end{align*}

where \begin{align*}r_1 = 2, r_2 = 7, \theta_1 = \frac{\pi}{2}, \theta_2 = \frac{3\pi}{2}\end{align*},

you can use the equation

\begin{align*}r_1r_2[\cos (\theta_1+\theta_2)+i \sin(\theta_1+\theta_2)]\end{align*}

and calculate:

\begin{align*}(2)(7)[\cos (\frac{\pi}{2}+\frac{3\pi}{2})+i \sin(\frac{\pi}{2}+\frac{3\pi}{2})]\end{align*}

This simplifies to:

\begin{align*}14[\cos(2\pi) + i\sin(2\pi)]\\ =14[1 + i0]\\ =14\\ \end{align*}

### Practice

Multiply each pair of complex numbers. If they are not in trigonometric form, change them before multiplying.

1. \begin{align*}3(\cos 32^\circ+i\sin 32^\circ)\cdot 2(\cos 15^\circ +i\sin 15^\circ )\end{align*}
2. \begin{align*}2(\cos 10^\circ+i\sin 10^\circ)\cdot 10(\cos 12^\circ +i\sin 12^\circ )\end{align*}
3. \begin{align*}4(\cos 45^\circ+i\sin 45^\circ)\cdot 8(\cos 62^\circ +i\sin 62^\circ )\end{align*}
4. \begin{align*}2(\cos 60^\circ+i\sin 60^\circ)\cdot \frac{1}{2}(\cos 34^\circ +i\sin 34^\circ )\end{align*}
5. \begin{align*}5(\cos 25^\circ+i\sin 25^\circ)\cdot 2(\cos 115^\circ +i\sin 115^\circ )\end{align*}
6. \begin{align*}-3(\cos 70^\circ+i\sin 70^\circ)\cdot 3(\cos 85^\circ +i\sin 85^\circ )\end{align*}
7. \begin{align*}7(\cos 85^\circ+i\sin 85^\circ)\cdot \sqrt{2}(\cos 40^\circ +i\sin 40^\circ )\end{align*}
8. \begin{align*}(3-2i)\cdot (1+i)\end{align*}
9. \begin{align*}(1-i)\cdot (1+i)\end{align*}
10. \begin{align*}(4-i)\cdot (3+2i)\end{align*}
11. \begin{align*}(1+i)\cdot (1+4i)\end{align*}
12. \begin{align*}(2+2i)\cdot (3+i)\end{align*}
13. \begin{align*}(1-3i)\cdot (2+i)\end{align*}
14. \begin{align*}(1-i)\cdot (1-i)\end{align*}
15. Can you multiply a pair of complex numbers in standard form without converting to trigonometric form?

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