# 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.

give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena