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Quotient Theorem

Simplified way to divide complex numbers.

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Quotient Theorem

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

Quotient Theorem

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 \begin{align*}r_1(\cos \theta_1 + i \sin \theta_1)\end{align*} and the second number be \begin{align*}r_2(\cos \theta_2 + i \sin \theta_2)\end{align*}. The product can then be simplified by use of five facts: the definition \begin{align*}i^2 = -1\end{align*}, the difference identity \begin{align*}\cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta)\end{align*}, the difference identity \begin{align*}\sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin (\alpha - \beta)\end{align*}, the Pythagorean identity, and the fact that the conjugate of \begin{align*}\cos \theta_2 + i \sin \theta_2\end{align*} is \begin{align*}\cos \theta_2 - i \sin \theta_2\end{align*}.

\begin{align*}& \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)]\end{align*}

In general:

\begin{align*}\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)]\end{align*}

Use this rule for the computation of two complex numbers divided by one another in the following problems. 

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

Express each number in polar form.

\begin{align*}& \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\end{align*}

\begin{align*}& \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\end{align*}

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

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

Find the quotient of the two complex numbers \begin{align*}28 \angle 35^\circ\end{align*} and \begin{align*}14 \angle 24^\circ\end{align*}

\begin{align*}& \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\end{align*}

\begin{align*} \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\end{align*}

Using the Quotient Theorem determine \begin{align*}\frac{1}{4cis \frac{\pi}{6}}\end{align*}

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

\begin{align*}1 \rightarrow x=1, y=0\end{align*}

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

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

Examples

Example 1

Earlier, you were given two complex numbers in polar form and was asked to divide them. 

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

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

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

Substituting values into this equation gives:

\begin{align*}\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}\\ \end{align*}

Example 2

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

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

\begin{align*}\frac{2 \angle 56^\circ}{7 \angle 113^\circ}=\frac{2}{7} \angle (56^\circ-113^\circ)=\frac{2}{7} \angle -57^\circ\end{align*}

Example 3

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

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


\begin{align*}\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)\end{align*}

Example 4

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

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

\begin{align*}\frac{2+3i}{-5+11i} \rightarrow \ \text{change each 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*}\frac{3.61}{12.08} \angle (56.31^\circ-114.44^\circ)=0.30 \angle - 58.13^\circ\end{align*}

Review

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

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

Review (Answers)

To see the Review answers, open this PDF file and look for section 6.11. 

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