# 10.5 Polar form of complex numbers  (Page 3/8)

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## 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.

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
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Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
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-x=7
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siame
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deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
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solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
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More example of algebra and trigo
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Navin