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Trigonometric Form of Complex Numbers

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Trigonometric Polar Form of Complex Numbers

You already know how to represent complex numbers in the complex plane using rectangular coordinates and you already know how to multiply and divide complex numbers.  Representing these points and performing these operations using trigonometric polar form will make your computations more efficient. 

What are the two ways to multiply the following complex numbers?

\left(1+\sqrt{3}i \right) \left(\sqrt{2}-\sqrt{2}i \right)

Watch This

http://www.youtube.com/watch?v=Zha7ZF8aVhU James Sousa: Trigonometric Form of Complex Numbers

Guidance

Any point represented in the complex plane as a+bi  can be represented in polar form just like any point in the rectangular coordinate system.  You will use the distance from the point to the origin as r  and the angle that the point makes as \theta

As you can see, the point a+bi  can also be represented as r \cdot \cos \theta+i \cdot r \cdot \sin \theta .  The trigonometric polar form can be abbreviated by factoring out the r  and noting the first letters:

r(\cos \theta+i \cdot \sin \theta) \rightarrow r \cdot \text{cis} \ \theta

The abbreviation r \cdot \text{cis} \ \theta  is read as “ r kiss theta.”  It allows you to represent a point as a radius and an angle.  One great benefit of this form is that it makes multiplying and dividing complex numbers extremely easy.  For example:

Let: z_1 = r_1 \cdot \text{cis} \ \theta_1, z_2=r_2 \cdot \text{cis} \ \theta_2 with r_2 \neq 0 .

Then:

z_1 \cdot z_2 &= r_1 \cdot r_2 \cdot \text{cis} \ (\theta_1 + \theta_2)\\z_1 \div z_2 &= \frac{r_1}{r_2} \cdot \text{cis} (\theta_1-\theta_2)

For basic problems, the amount of work required to compute products and quotients for complex numbers given in either form is roughly equivalent.  For more challenging questions, trigonometric polar form becomes significantly advantageous.

Example A

Convert the following complex number from rectangular form to trigonometric polar form.

1 - \sqrt{3}i

Solution:   The radius is the absolute value of the number.

r^2=1^2+ \left(-\sqrt{3}\right)^2 \rightarrow r=2

The angle can be found with basic trig and the knowledge that the opposite side is always the imaginary component and the adjacent side is always the real component.

\tan \theta=-\frac{\sqrt{3}}{1} \rightarrow \theta=60^\circ

Thus the trigonometric form is  2 \ \text{cis}\ 60^\circ .

Example B

Convert the following complex number from trigonometric polar form to rectangular form.

4 \ \text{cis} \left(\frac{3 \pi}{4}\right)

Solution: 4 \ \text{cis} \ \left(\frac{3 \pi}{4}\right)=4 \left(\cos \left(\frac{3 \pi}{4}\right)+i \cdot \sin \left(\frac{3 \pi}{4}\right)\right)=4 \left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=-2 \sqrt{2}+2 \sqrt{2}i

Example C

Divide the following complex numbers.

\frac{4 \ \text{cis} \ 32^\circ}{2 \ \text{cis} \ 2^\circ}

Solution: \frac{4 \ \text{cis} \ 32^\circ}{2 \ \text{cis} \ 2^\circ}=\frac{4}{2} \ \text{cis} \ (32^\circ-2^\circ)=2 \ \text{cis} \ (30^\circ)

Concept Problem Revisited

In rectangular coordinates:

\left(1+\sqrt{3}i\right)\left(\sqrt{2}-\sqrt{2}i\right)=\sqrt{2}-\sqrt{2}i+\sqrt{6}i+\sqrt{6}

In trigonometric polar coordinates, 1+\sqrt{3}i=2 \ \text{cis} \ 60^\circ  and \sqrt{2}-\sqrt{2}i=2 \ \text{cis}-45^\circ .  Therefore:

\left(1+\sqrt{3}i\right) \left(\sqrt{2}-\sqrt{2}i \right)=2 \ \text{cis} \ 60^\circ \cdot 2 \ \text{cis} - 45^\circ=4 \ \text{cis} \ 105^\circ

Vocabulary

Trigonometric polar form of a complex number describes the location of a point on the complex plane using the angle and the radius of the point. 

The abbreviation r \cdot \text{cis} \ \theta  stands for r \cdot (\cos \theta+i \cdot \sin \theta)  and is how trigonometric polar form is typically observed. 

Guided Practice

1. Translate the following complex number from trigonometric polar form to rectangular form.

5 \ \text{cis} \ 270^\circ

2. Translate the following complex number from rectangular form into trigonometric polar form.

8

3. Multiply the following complex numbers in trigonometric polar form.

4 \ \text{cis} \ 34^\circ \cdot 5 \ \text{cis} \ 16^\circ \cdot \frac{1}{2} \ \text{cis} \ 100^\circ

Answers:   

1. 5 \ \text{cis} \ 270^\circ =5 (\cos 270^\circ+i \cdot \sin 270^\circ)=5 (0-i)=-5 i

2. 8=8 \ \text{cis} \ 0^\circ

3. & 4 \ \text{cis} \ 34^\circ \cdot 5 \ \text{cis} \ 16^\circ \cdot \frac{1}{2} \ \text{cis} \ 100^\circ\\& =4 \cdot 5 \cdot \frac{1}{2} \cdot \text{cis} \ (34^\circ+16^\circ+100^\circ)=10 \ \text{cis} \ 150

Note how much easier it is to do products and quotients in trigonometric polar form. 

Practice

Translate the following complex numbers from trigonometric polar form to rectangular form. 

1. 5 \ \text{cis} \ 270^\circ

2.  2 \ \text{cis} \ 30^\circ

3.  -4 \ \text{cis} \ \frac{\pi}{4}

4.  6 \ \text{cis} \ \frac{\pi}{3}

5.  2 \ \text{cis} \ \frac{5 \pi}{2}

Translate the following complex numbers from rectangular form into trigonometric polar form.

6. 2-i

7.  5+12i

8.  6i+8

9.  i

Complete the following calculations and simplify.

10. 2 \ \text{cis} \ 22^\circ \cdot \frac{1}{5} \ \text{cis} \ 15^\circ \cdot 3 \ \text{cis} \ 95^\circ

11. 9 \ \text{cis} \ 98^\circ \div 3 \ \text{cis} \ 12^\circ

12. 15 \ \text{cis} \ \frac{\pi}{4} \cdot 2 \ \text{cis} \ \frac{\pi}{6}

13. -2 \ \text{cis} \ \frac{2 \pi}{3} \div 15 \ \text{cis} \ \frac{7 \pi}{6}

Let z_1=r_1 \cdot \text{cis} \ \theta_1  and  z_2=r_2 \cdot \text{cis} \ \theta_2 with r_2 \neq 0 .

14.  Use the trigonometric sum and difference identities to prove that z_1 \cdot z_2=r_1 \cdot r_2 \cdot \text{cis} \ (\theta_1+\theta_2) .

15.  Use the trigonometric sum and difference identities to prove that z_1 \div z_2=\frac{r_1}{r_2} \cdot \text{cis} \ (\theta_1-\theta_2) .

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