8.5 Polar form of complex numbers  (Page 3/8)

 Page 3 / 8

Products of complex numbers in polar form

If $\text{\hspace{0.17em}}{z}_{1}={r}_{1}\left(\mathrm{cos}\text{\hspace{0.17em}}{\theta }_{1}+i\mathrm{sin}\text{\hspace{0.17em}}{\theta }_{1}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{z}_{2}={r}_{2}\left(\mathrm{cos}\text{\hspace{0.17em}}{\theta }_{2}+i\mathrm{sin}\text{\hspace{0.17em}}{\theta }_{2}\right),$ then the product of these numbers is given as:

$\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}={r}_{1}{r}_{2}\text{cis}\left({\theta }_{1}+{\theta }_{2}\right)\hfill \end{array}\hfill \end{array}$

Notice that the product calls for multiplying the moduli and adding the angles.

Finding the product of two complex numbers in polar form

Find the product of $\text{\hspace{0.17em}}{z}_{1}{z}_{2},\text{\hspace{0.17em}}$ given $\text{\hspace{0.17em}}{z}_{1}=4\left(\mathrm{cos}\left(80°\right)+i\mathrm{sin}\left(80°\right)\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{z}_{2}=2\left(\mathrm{cos}\left(145°\right)+i\mathrm{sin}\left(145°\right)\right).$

Follow the formula

$\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\mathrm{cos}\left(80°+145°\right)+i\mathrm{sin}\left(80°+145°\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\mathrm{cos}\left(225°\right)+i\mathrm{sin}\left(225°\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\mathrm{cos}\left(\frac{5\pi }{4}\right)+i\mathrm{sin}\left(\frac{5\pi }{4}\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\hfill \end{array}$

Finding quotients of complex numbers in polar form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

Quotients of complex numbers in polar form

If $\text{\hspace{0.17em}}{z}_{1}={r}_{1}\left(\mathrm{cos}\text{\hspace{0.17em}}{\theta }_{1}+i\mathrm{sin}\text{\hspace{0.17em}}{\theta }_{1}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{z}_{2}={r}_{2}\left(\mathrm{cos}\text{\hspace{0.17em}}{\theta }_{2}+i\mathrm{sin}\text{\hspace{0.17em}}{\theta }_{2}\right),$ then the quotient of these numbers is

$\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}-{\theta }_{2}\right)\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}{z}_{2}\ne 0\\ \frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\text{cis}\left({\theta }_{1}-{\theta }_{2}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{z}_{2}\ne 0\text{\hspace{0.17em}}\end{array}$

Notice that the moduli are divided, and the angles are subtracted.

Given two complex numbers in polar form, find the quotient.

1. Divide $\text{\hspace{0.17em}}\frac{{r}_{1}}{{r}_{2}}.$
2. Find $\text{\hspace{0.17em}}{\theta }_{1}-{\theta }_{2}.$
3. Substitute the results into the formula: $\text{\hspace{0.17em}}z=r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right).\text{\hspace{0.17em}}$ Replace $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\frac{{r}_{1}}{{r}_{2}},\text{\hspace{0.17em}}$ and replace $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}{\theta }_{1}-{\theta }_{2}.$
4. Calculate the new trigonometric expressions and multiply through by $\text{\hspace{0.17em}}r.$

Finding the quotient of two complex numbers

Find the quotient of $\text{\hspace{0.17em}}{z}_{1}=2\left(\mathrm{cos}\left(213°\right)+i\mathrm{sin}\left(213°\right)\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{z}_{2}=4\left(\mathrm{cos}\left(33°\right)+i\mathrm{sin}\left(33°\right)\right).$

Using the formula, we have

$\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\mathrm{cos}\left(213°-33°\right)+i\mathrm{sin}\left(213°-33°\right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\mathrm{cos}\left(180°\right)+i\mathrm{sin}\left(180°\right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\hfill \end{array}$

Find the product and the quotient of $\text{\hspace{0.17em}}{z}_{1}=2\sqrt{3}\left(\mathrm{cos}\left(150°\right)+i\mathrm{sin}\left(150°\right)\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{z}_{2}=2\left(\mathrm{cos}\left(30°\right)+i\mathrm{sin}\left(30°\right)\right).$

$\text{\hspace{0.17em}}{z}_{1}{z}_{2}=-4\sqrt{3};\frac{{z}_{1}}{{z}_{2}}=-\frac{\sqrt{3}}{2}+\frac{3}{2}i\text{\hspace{0.17em}}$

Finding powers of complex numbers in polar form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem    . It states that, for a positive integer $\text{\hspace{0.17em}}n,{z}^{n}\text{\hspace{0.17em}}$ is found by raising the modulus to the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ power and multiplying the argument by $\text{\hspace{0.17em}}n.\text{\hspace{0.17em}}$ It is the standard method used in modern mathematics.

De moivre’s theorem

If $\text{\hspace{0.17em}}z=r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\text{\hspace{0.17em}}$ is a complex number, then

$\begin{array}{l}{z}^{n}={r}^{n}\left[\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}$

where $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is a positive integer.

Evaluating an expression using de moivre’s theorem

Evaluate the expression $\text{\hspace{0.17em}}{\left(1+i\right)}^{5}\text{\hspace{0.17em}}$ using De Moivre’s Theorem.

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write $\text{\hspace{0.17em}}\left(1+i\right)\text{\hspace{0.17em}}$ in polar form. Let us find $\text{\hspace{0.17em}}r.$

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}\hfill \\ r=\sqrt{2}\hfill \end{array}$

Then we find $\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Using the formula $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{y}{x}\text{\hspace{0.17em}}$ gives

$\begin{array}{l}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{1}{1}\hfill \\ \mathrm{tan}\text{\hspace{0.17em}}\theta =1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =\frac{\pi }{4}\hfill \end{array}$

Use De Moivre’s Theorem to evaluate the expression.

$\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\mathrm{cos}\left(5\cdot \frac{\pi }{4}\right)+i\mathrm{sin}\left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(1+i\right)}^{5}=4\sqrt{2}\left[\mathrm{cos}\left(\frac{5\pi }{4}\right)+i\mathrm{sin}\left(\frac{5\pi }{4}\right)\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(1+i\right)}^{5}=-4-4i\hfill \end{array}$

Finding roots of complex numbers in polar form

To find the n th root of a complex number in polar form, we use the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ Root Theorem or De Moivre’s Theorem    and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ roots of complex numbers in polar form.

The n Th root theorem

To find the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ root of a complex number in polar form, use the formula given as

${z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\mathrm{cos}\left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\mathrm{sin}\left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right]$

where $\text{\hspace{0.17em}}k=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}n-1.\text{\hspace{0.17em}}$ We add $\text{\hspace{0.17em}}\frac{2k\pi }{n}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\frac{\theta }{n}\text{\hspace{0.17em}}$ in order to obtain the periodic roots.

Questions & Answers

how can are find the domain and range of a relations
austin Reply
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply
i want to sure my answer of the exercise
meena Reply
what is the diameter of(x-2)²+(y-3)²=25
Den Reply
how to solve the Identity ?
Barcenas Reply
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
Shakeena Reply
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
Rhudy Reply
what is a complex number used for?
Drew Reply
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim

Read also:

Get the best Precalculus course in your pocket!

Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

 By By By By Mldelatte By Qqq Qqq