<meta http-equiv="refresh" content="1; url=/nojavascript/"> Quotient Theorem ( Read ) | Trigonometry | CK-12 Foundation
Dismiss
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Quotient Theorem

%
Best Score
Practice Quotient Theorem
Practice
Best Score
%
Practice Now

Quotient Theorem

Suppose you are given two complex numbers in polar form, such as 2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) and 3(\cos \pi+i \sin \pi) and asked to divide them. Can you do this? How long will it take you?

By the end of this Concept, you'll know how to divide complex numbers using the Quotient Theorem.

Watch This

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

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

Guidance

Division of complex numbers in polar form is similar to the division of complex numbers in standard form. However, to determine a general rule for division, the denominator must be rationalized by multiplying the fraction by the complex conjugate of the denominator. In addition, the trigonometric functions must be simplified by applying the sum/difference identities for cosine and sine as well as one of the Pythagorean identities. To obtain a general rule for the division of complex numbers in polar from, let the first number be r_1(\cos \theta_1 + i \sin \theta_1) and the second number be r_2(\cos \theta_2 + i \sin \theta_2) . The product can then be simplified by use of five facts: the definition i^2 = -1 , the difference identity \cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta) , the difference identity \sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin (\alpha - \beta) , the Pythagorean identity, and the fact that the conjugate of \cos \theta_2 + i \sin \theta_2 is \cos \theta_2 - i \sin \theta_2 .

& \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}\\& \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)} \cdot \frac{(\cos \theta_2-i \sin \theta_2)}{(\cos \theta_2-i \sin \theta_2)}\\& \frac{r_1}{r_2} \cdot \frac{\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}{\cos^2 \theta_2-i^2 \sin^2 \theta_2}\\& \frac{r_1}{r_2} \cdot \frac{(\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)}{\cos^2 \theta_2 + \sin^2 \theta_2}\\& \frac{r_1}{r_2}[\cos (\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]

In general:

\frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]

We can use this rule for the computation of two complex numbers divided by one another.

Example A

Find the quotient of (\sqrt{3}-i) \div (2- i2\sqrt{3})

Solution: Express each number in polar form.

& \sqrt{3}-i && 2-i2\sqrt{3}\\& r_1=\sqrt{x^2+y^2} && r_2=\sqrt{x^2+y^2}\\& r_1=\sqrt{(\sqrt{3})^2+(-1)^2} && r_2  = \sqrt{(2)^2+(-2\sqrt{3})^2}\\& r_1=\sqrt{4}=2 && r_2=\sqrt{16}=4

& \frac{r_1}{r_2}=.5\\& \theta_1=\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) && \theta_2=\tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) && \theta=\theta_1-\theta_2\\& \theta_1=5.75959 \ rad. && \theta_2=5.23599 \ rad. && \theta=5.75959-5.23599\\&&&&& \theta=0.5236

Now, plug in what we found to the Quotient Theorem.

\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]=.5(\cos 0.5236+i \sin 0.5236)

Example B

Find the quotient of the two complex numbers 28 \angle 35^\circ and 14 \angle 24^\circ

Solution:

& \text{For} \ 28 \ \angle 35^\circ && \text{For} \ 14 \ \angle 24^\circ && \frac{r_1}{r_2}=\frac{28}{14}=2\\& r_1=28 && r_2=14 && \theta=\theta_1-\theta_2\\& \theta_1=35^\circ && \theta_2=24^\circ && \theta=35^\circ-24^\circ=11^\circ

 \frac{r_1 \angle \theta_1}{r_2 \angle \theta_2} &= \frac{r_1}{r_2} \angle (\theta_1-\theta_2)\\&=2 \angle 11^\circ

Example C

Using the Quotient Theorem determine \frac{1}{4cis \frac{\pi}{6}}

Solution:

Even though 1 is not a complex number, we can still change it to polar form.

1 \rightarrow x=1, y=0

r &= \sqrt{1^2+0^2}=1 \qquad \\\tan \theta &= \frac{0}{1}=0 \rightarrow \theta = 0^\circ

\text{So}, \frac{1}{4cis\frac{\pi}{6}}=\frac{1cis0}{4cis\frac{\pi}{6}}=\frac{1}{4}cis \left(0-\frac{\pi}{6}\right)=\frac{1}{4}cis\left(-\frac{\pi}{6}\right).

Vocabulary

Quotient Theorem: The quotient theorem is a theorem showing a simplified way to divide complex numbers.

Guided Practice

1. Divide the following complex numbers. If they are not in polar form, change them before dividing.

\frac{2 \angle 56^\circ}{7 \angle 113^\circ}

2. Divide the following complex numbers. If they are not in polar form, change them before dividing.

\frac{10 \left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}

3. Divide the following complex numbers. If they are not in polar form, change them before dividing.

\frac{2+3i}{-5+11i}

Solutions:

1. \frac{2 \angle 56^\circ}{7 \angle 113^\circ}=\frac{2}{7} \angle (56^\circ-113^\circ)=\frac{2}{7} \angle -57^\circ

2. \frac{10\left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}&=2 \left(\cos \left(\frac{5\pi}{3}-\pi\right)+i \sin \left(\frac{5\pi}{3}-\pi\right)\right)\\ &=2\left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)

3. \frac{2+3i}{-5+11i} \rightarrow \ \text{change each to polar}. & 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 \frac{3.61}{12.08} \angle (56.31^\circ-114.44^\circ)=0.30 \angle - 58.13^\circ

Concept Problem Solution

You know that the 2 numbers to divide are 2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) and 3(\cos \pi+i \sin \pi) .

If you consider r_1 = 2,r_2 = 3, \theta_1 = \frac{\pi}{3}, \theta_2 = \pi , you can use the formula:

\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]

Substituting values into this equation gives:

\frac{2}{3}[\cos(\frac{\pi}{3}-\pi)+i \sin(\frac{\pi}{3}-\pi)]\\=\frac{2}{3}[\cos \left( -\frac{2\pi}{3} \right) + i \sin \left( -\frac{2\pi}{3} \right)\\=\frac{2}{3}\left( \left( -\frac{1}{2} \right) + i \left( -\frac{\sqrt{3}}{2} \right) \right)\\=-\frac{1}{3} - i\frac{\sqrt{3}}{3}\\

Practice

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

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

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...

Original text