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anonymous
 one year ago
Thanks for helping!
I need to calculate wz using the De Moivre's theorem. z=sqrt(3)i and w=22i. I need to rewrite it in polar form. Thank You!
anonymous
 one year ago
Thanks for helping! I need to calculate wz using the De Moivre's theorem. z=sqrt(3)i and w=22i. I need to rewrite it in polar form. Thank You!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Ashleyisakitty @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate @wio @sammixboo

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3In 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
 one year ago
Best ResponseYou've already chosen the best response.3Multiply the radial lengths, add the angles. So we gotta do some work to get it into polar form first, ya? :d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02*2sqrt(2)(cos(pi/6+5pi/4)+isin(pi/6+5pi/4))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I just calculate that?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Mmm ok ya, looks good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for theta I would put 13pi/12?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3and 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
 one year ago
Best ResponseYou've already chosen the best response.0Yup, but my teacher wants full form unfortunately :(

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Yes that looks correct! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome! Thanks for the help! Can I have help with one more question really quickly?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0z^2 so (sqrt(3)i)^2. I think it is 3+i, but I have a feeling I did it wrong.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm z=2cis\left(\frac{\pi}{6}\right)\]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
 one year ago
Best ResponseYou've already chosen the best response.3Oh I guess NOT converting to polar is probably better :) lol is that what you did?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh crap, sorry, I forgot to say I needed to write in a+bi form

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm z=(\sqrt3i)\]\[\Large\rm z^2=(\sqrt3i)(\sqrt3i)\]\[\Large\rm z^2=3\sqrt3 i\sqrt3 i+i^2\]Something like that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that all I need to do? I thought there was some formula or something. I did (a)^+(b+^2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3No, that's when you're multiplying complex `conjugates`.\[\Large\rm (a+bi)(abi)=a^2+b^2\]This is not the same as squaring something! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm (a+bi)(abi)\ne(a+bi)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, so I would leave my results as i^2sqrt(3)isqrt(3)i+3?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3No no :) We can simplify further. I just wanted to show the FOIL process

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So we have some liketerms in the middle which can be combined. And what do you get when you square i?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i^2? or if you mean root, i

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm \sqrt{1}\cdot\sqrt{1}=1\]Yah that looks better :) I made a silly mistake lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm z^2=32\sqrt3 i1\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Simplify furtherrrrr :U

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3maybe we write the a first, a+(b)i = 2 + (2sqrt3)i ok looks good! \c:/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, I presume it's the exact same if I were to do w^4? Thanks for the help, I learned a lot :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Expanding 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm w=2\sqrt{2}cis(5\pi/4)\]\[\Large\rm w^4=(2\sqrt{2})^4cis(5\pi/4)^4\]\[\Large\rm w^4=(2\sqrt{2})^4cis(4\cdot 5\pi/4)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3ah sorry i ran away for a bit hehe

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So if THAT direction is 3, which direction is positive 3? maybe you can label it on the graph.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434425967838:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434426009450:dwThink about your regular xy plane. if i go 3 in this direction,

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434426045344:dwThen the opposite (negative) of that is to go 3 in the other direction, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434426087433:dwSo we want to go 3 in the opposite direction

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Mmm ya that sounds right: pi/3 + pi = 4pi/3 or 2pi/3 would work also. the positive angle seems nicer though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, once again thanks!
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