Question

Thanks for helping!

I need to calculate wz using the De Moivre's theorem. z=sqrt(3)-i and w=-2-2i. I need to rewrite it in polar form. Thank You!

Answer 1

@Ashleyisakitty @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate @wio @sammixboo

Answer 2

In general when we have:\[\Large\rm z_1=r_1(\cos \alpha+i \sin \alpha)\]\[\Large\rm z_2=r_2(\cos \beta+i \sin \beta)\]Then multiplying will give us:\[\Large\rm z_1\cdot z_2=r_1 \cdot r_2 \left(\cos(\alpha + \beta)+i \sin(\alpha + \beta)\right)\]

Answer 3

Multiply the radial lengths,
add the angles.
So we gotta do some work to get it into polar form first, ya? :d

Answer 4

already done :D

Answer 5

2*2sqrt(2)(cos(-pi/6+5pi/4)+isin(-pi/6+5pi/4))

Answer 6

Then I just calculate that?

Answer 7

Mmm ok ya, looks good!

Answer 8

so for theta I would put 13pi/12?

Answer 9

and r is 4sqrt(2)

Answer 10

and you can use CIS if you wanna be lazy.\[\Large\rm z=2 cis\left(-\frac{\pi}{6}\right),\qquad w=2\sqrt2 cis\left(\frac{5\pi}{4}\right)\]I like to do that sometimes hehe

Answer 11

Yup, but my teacher wants full form unfortunately :(

Answer 12

Yes that looks correct! :)

Answer 13

awesome! Thanks for the help! Can I have help with one more question really quickly?

Answer 14

sure

Answer 15

z^2 so (sqrt(3)-i)^2. I think it is 3+i, but I have a feeling I did it wrong.

Answer 16

\[\Large\rm z=2cis\left(-\frac{\pi}{6}\right)\]Then,\[\Large\rm z^2=2^2cis\left(-\frac{\pi}{6}\right)^2\]De'Moivre's let's us bring the exponent inside,\[\Large\rm z^2=4cis\left(-\frac{2\pi}{6}\right)\]
So that angle is -pi/3 ya?
So our imaginary part should be negative, we're still in quadrant 4.

Answer 17

Oh I guess NOT converting to polar is probably better :) lol
is that what you did?

Answer 18

yup

Answer 19

oh crap, sorry, I forgot to say I needed to write in a+bi form

Answer 20

my bad

Answer 21

\[\Large\rm z=(\sqrt3-i)\]\[\Large\rm z^2=(\sqrt3-i)(\sqrt3-i)\]\[\Large\rm z^2=3-\sqrt3 i-\sqrt3 i+i^2\]Something like that?

Answer 22

Is that all I need to do? I thought there was some formula or something. I did (a)^+(b+^2

Answer 23

(b)^2*

Answer 24

No, that's when you're multiplying complex `conjugates`.\[\Large\rm (a+bi)(a-bi)=a^2+b^2\]This is not the same as squaring something! :)

Answer 25

\[\Large\rm (a+bi)(a-bi)\ne(a+bi)^2\]

Answer 26

Oh, so I would leave my results as i^2-sqrt(3)i-sqrt(3)i+3?

Answer 27

No no :) We can simplify further.
I just wanted to show the FOIL process

Answer 28

So we have some like-terms in the middle which can be combined.
And what do you get when you square i?

Answer 29

i^2? or if you mean root, i

Answer 30

-2sqrt(3)i

Answer 31

woops!

Answer 32

-1?

Answer 33

\[\Large\rm \sqrt{-1}\cdot\sqrt{-1}=-1\]Yah that looks better :)
I made a silly mistake lol

Answer 34

\[\Large\rm z^2=3-2\sqrt3 i-1\]

Answer 35

Simplify furtherrrrr :U

Answer 36

-2sqrt(3)i+2

Answer 37

maybe we write the a first,
a+(b)i = 2 + (-2sqrt3)i

ok looks good! \c:/

Answer 38

Awesome, I presume it's the exact same if I were to do w^4? Thanks for the help, I learned a lot :D

Answer 39

Expanding a 4th degree is really a pain in the butt,
lot of brackets to multiply out.

When you get above a 3rd power, you want to start doing it the other way:
Change to polar,
apply De'Moivre's Theorem.

Answer 40

\[\Large\rm w=2\sqrt{2}cis(5\pi/4)\]\[\Large\rm w^4=(2\sqrt{2})^4cis(5\pi/4)^4\]\[\Large\rm w^4=(2\sqrt{2})^4cis(4\cdot 5\pi/4)\]

Answer 41

64(cis(5pi))?

Answer 42

I apologize if I'm taking too much of your time, but may I ask a quick question?

If I have to write the coordinates of (-3, pi/3) as a positive value (as in radius), would I write (3, pi/3)? Thanks!

Answer 43

ah sorry i ran away for a bit hehe

Answer 44

That is fine :)