Mathematics
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OpenStudy (anonymous):
Convert the complex number 2 - 2root3 into its polar representation. ??
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OpenStudy (anonymous):
@SolomonZelman @perl @Kainui
OpenStudy (anonymous):
@OOOPS @Oooky
OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
\((2, -2\sqrt3)\), right?
OpenStudy (anonymous):
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OpenStudy (anonymous):
OpenStudy (anonymous):
first off, x =2, y = -2root3, hence x^2+y^2 =r^2 =?
OpenStudy (anonymous):
I guide you how to find it out, I don't do your homework
OpenStudy (anonymous):
answer me step by step, you can get it, it is not hard
x^2 +y^2 =?
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OpenStudy (anonymous):
right ok so x^2 is 4 and
y^2 = 12?
OpenStudy (anonymous):
@Oooky on test, we don't have math tools
OpenStudy (anonymous):
so r = 4?
OpenStudy (anonymous):
yup
OpenStudy (anonymous):
It was for future reference :P
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OpenStudy (anonymous):
now, take r out by factor (2, -2root3) what do you get?
OpenStudy (anonymous):
ok, may be this is hard for you, I guide you this :)
OpenStudy (anonymous):
yeah how do i do that haha
OpenStudy (anonymous):
good luck :)))
OpenStudy (anonymous):
for x =2 , do you agree that 2 = 4*\(\dfrac{2}{4}\)
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OpenStudy (anonymous):
hence if I take 4 out, then the rest is 1/2, right?
OpenStudy (anonymous):
yup i got that
OpenStudy (anonymous):
do the same with y, we have y = \(-2\sqrt 3= 4*-2\sqrt3/4 \) if I take 4 out, I have the rest is \(-\dfrac{\sqrt 3}{2}\), right?
OpenStudy (anonymous):
so far, I have \(4(\dfrac{1}{2}, -\dfrac{\sqrt3}{2}) \) got me so far?
I just take 4 out from the original (x, y)
OpenStudy (anonymous):
yeah I follow that
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OpenStudy (anonymous):
cos ??= 1/2?
OpenStudy (anonymous):
pi/3?
OpenStudy (anonymous):
oh, there is an easier way to do, hihihi... but by mistake, I go on this way. go ahead or go other way?? hahahah....
but the step to get r is perfect,
OpenStudy (anonymous):
either way haha
OpenStudy (anonymous):
ok, \(tan \theta =\dfrac{y}{x}\), just put the original x, y in to get tan
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OpenStudy (anonymous):
I get tan theta = root 3
OpenStudy (anonymous):
-root 3, right? not root 3
OpenStudy (anonymous):
that's pi/3
OpenStudy (anonymous):
oh right so -pi/3
OpenStudy (anonymous):
positive one, please
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OpenStudy (anonymous):
so is it A?
OpenStudy (anonymous):
|dw:1419812456038:dw|
OpenStudy (anonymous):
\[\sqrt[4]{3} \cos(60) = i \sin(60)\]
OpenStudy (anonymous):
|dw:1419812490452:dw|