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!

- anonymous

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- anonymous

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

- zepdrix

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:\[\Large\rm z_1\cdot z_2=r_1 \cdot r_2 \left(\cos(\alpha + \beta)+i \sin(\alpha + \beta)\right)\]

- zepdrix

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

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## More answers

- anonymous

already done :D

- anonymous

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

- anonymous

Then I just calculate that?

- zepdrix

Mmm ok ya, looks good!

- anonymous

so for theta I would put 13pi/12?

- anonymous

and r is 4sqrt(2)

- zepdrix

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

- anonymous

Yup, but my teacher wants full form unfortunately :(

- zepdrix

Yes that looks correct! :)

- anonymous

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

- zepdrix

sure

- anonymous

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

- zepdrix

\[\Large\rm z=2cis\left(-\frac{\pi}{6}\right)\]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.

- zepdrix

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

- anonymous

yup

- anonymous

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

- anonymous

my bad

- zepdrix

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

- anonymous

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

- anonymous

(b)^2*

- zepdrix

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! :)

- zepdrix

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

- anonymous

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

- zepdrix

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

- zepdrix

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

- anonymous

i^2? or if you mean root, i

- anonymous

-2sqrt(3)i

- zepdrix

woops!

- anonymous

-1?

- zepdrix

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

- zepdrix

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

- zepdrix

Simplify furtherrrrr :U

- anonymous

-2sqrt(3)i+2

- zepdrix

maybe we write the a first,
a+(b)i = 2 + (-2sqrt3)i
ok looks good! \c:/

- anonymous

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

- zepdrix

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.

- zepdrix

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

- anonymous

64(cis(5pi))?

- anonymous

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!

- zepdrix

ah sorry i ran away for a bit hehe

- anonymous

That is fine :)

- zepdrix

|dw:1434425783548:dw|

- zepdrix

So if THAT direction is -3,
which direction is positive 3?
maybe you can label it on the graph.

- anonymous

|dw:1434425967838:dw|

- zepdrix

|dw:1434426009450:dw|Think about your regular x-y plane.
if i go 3 in this direction,

- zepdrix

|dw:1434426045344:dw|Then the opposite (negative) of that is to go 3 in the other direction, ya?

- anonymous

yea

- zepdrix

|dw:1434426087433:dw|So we want to go 3 in the opposite direction

- anonymous

Oh, gotcha

- anonymous

4pi/3?

- zepdrix

|dw:1434426162247:dw|

- zepdrix

Mmm ya that sounds right: pi/3 + pi = 4pi/3
or -2pi/3 would work also.
the positive angle seems nicer though

- anonymous

Awesome, once again thanks!

- zepdrix

np

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