# 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 $\text{\hspace{0.17em}}{z}_{2}=2\left(\mathrm{cos}\left(145°\right)+i\mathrm{sin}\left(145°\right)\right).$

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

## 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.

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
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?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert