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!

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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!

Mathematics
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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:\[\Large\rm z_1\cdot z_2=r_1 \cdot r_2 \left(\cos(\alpha + \beta)+i \sin(\alpha + \beta)\right)\]
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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already done :D
2*2sqrt(2)(cos(-pi/6+5pi/4)+isin(-pi/6+5pi/4))
Then I just calculate that?
Mmm ok ya, looks good!
so for theta I would put 13pi/12?
and r is 4sqrt(2)
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
Yup, but my teacher wants full form unfortunately :(
Yes that looks correct! :)
awesome! Thanks for the help! Can I have help with one more question really quickly?
sure
z^2 so (sqrt(3)-i)^2. I think it is 3+i, but I have a feeling I did it wrong.
\[\Large\rm z=2cis\left(-\frac{\pi}{6}\right)\]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.
Oh I guess NOT converting to polar is probably better :) lol is that what you did?
yup
oh crap, sorry, I forgot to say I needed to write in a+bi form
my bad
\[\Large\rm z=(\sqrt3-i)\]\[\Large\rm z^2=(\sqrt3-i)(\sqrt3-i)\]\[\Large\rm z^2=3-\sqrt3 i-\sqrt3 i+i^2\]Something like that?
Is that all I need to do? I thought there was some formula or something. I did (a)^+(b+^2
(b)^2*
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! :)
\[\Large\rm (a+bi)(a-bi)\ne(a+bi)^2\]
Oh, so I would leave my results as i^2-sqrt(3)i-sqrt(3)i+3?
No no :) We can simplify further. I just wanted to show the FOIL process
So we have some like-terms in the middle which can be combined. And what do you get when you square i?
i^2? or if you mean root, i
-2sqrt(3)i
woops!
-1?
\[\Large\rm \sqrt{-1}\cdot\sqrt{-1}=-1\]Yah that looks better :) I made a silly mistake lol
\[\Large\rm z^2=3-2\sqrt3 i-1\]
Simplify furtherrrrr :U
-2sqrt(3)i+2
maybe we write the a first, a+(b)i = 2 + (-2sqrt3)i ok looks good! \c:/
Awesome, I presume it's the exact same if I were to do w^4? Thanks for the help, I learned a lot :D
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.
\[\Large\rm w=2\sqrt{2}cis(5\pi/4)\]\[\Large\rm w^4=(2\sqrt{2})^4cis(5\pi/4)^4\]\[\Large\rm w^4=(2\sqrt{2})^4cis(4\cdot 5\pi/4)\]
64(cis(5pi))?
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!
ah sorry i ran away for a bit hehe
That is fine :)
|dw:1434425783548:dw|
So if THAT direction is -3, which direction is positive 3? maybe you can label it on the graph.
|dw:1434425967838:dw|
|dw:1434426009450:dw|Think about your regular x-y plane. if i go 3 in this direction,
|dw:1434426045344:dw|Then the opposite (negative) of that is to go 3 in the other direction, ya?
yea
|dw:1434426087433:dw|So we want to go 3 in the opposite direction
Oh, gotcha
4pi/3?
|dw:1434426162247:dw|
Mmm ya that sounds right: pi/3 + pi = 4pi/3 or -2pi/3 would work also. the positive angle seems nicer though
Awesome, once again thanks!
np

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