anonymous
  • anonymous
Which expression is a cube root of -1 + i sqrt(3)? A. cubert(2) (cos(120degrees) + i sin(120 degrees)) B. cubert(2) (cos(40 degrees) + i sin(40 degrees)) C. cubert(2) (cos(280 degrees) + i sin(280 degrees)) D. cubert(2) (cos(320 degrees) + i sin(320 degrees)) ***My Answer: D*** Please Help!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[-1+\sqrt{3}=r \left( \cos \theta+\iota \sin \theta \right)\] \[r \cos \theta =-1,r \sin \theta =\sqrt{3}\] square and add and find r divide find theta
anonymous
  • anonymous
So I would do r = -1-cos(theta), r = 3sin^2(theta)?
anonymous
  • anonymous
\[r^2\cos ^2\theta =1,r^2\sin ^2\theta=3\] \[r^2\left( \cos ^2\theta+\sin ^2\theta \right)=1+3=4\] r=2

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anonymous
  • anonymous
Oh ok
anonymous
  • anonymous
r is always positive ,so \[\cos~ \theta~ is~ negative~and~\sin \theta~is~positive.\] so angle lies in second quadrant. now divide to find theta
anonymous
  • anonymous
Would \[\theta = 120\] be the angle?
anonymous
  • anonymous
no guess ,show me your calculations.
anonymous
  • anonymous
Ok \[r ^{2}(\cos ^{2}\theta + \sin ^{2}\theta)\]\[2 (\cos ^{2}\theta + \sin ^{2}\theta)\] \[-1 + i \sqrt{3}\] \[\theta = \frac{ \sqrt{3} }{ -1}\]\[\theta = -\frac{ \pi }{ 3}\]
anonymous
  • anonymous
That is all I know of. Not exactly sure if it is correct. Sorry
anonymous
  • anonymous
\[\frac{ r \sin \theta }{ r \cos \theta }=\frac{ \sqrt{3} }{ -1 }=-\sqrt{3}=-\tan 60=\tan \left( 180-60\right) =\tan 120\]
anonymous
  • anonymous
\[\tan \theta=\tan 120,\theta=120\]
anonymous
  • anonymous
Oh wow I was far off but at the same time I kinda sorta knew what I was doing. Haha. I answer would now be A I'm assuming. Thank you for taking the time to explain this to me. I desperately needed the help.
anonymous
  • anonymous
\[-1+i \sqrt{3}=2\left( \cos 120+i \sin 120 \right)\]
anonymous
  • anonymous
yw
anonymous
  • anonymous
Oh wait the answer would be B
anonymous
  • anonymous
oh sorry you wanted cube root ,i have not noticed
anonymous
  • anonymous
It's ok :) I already hit the submit button haha
anonymous
  • anonymous
Wouldn't you just divide 120/3 to get 40?
anonymous
  • anonymous
\[1+i \sqrt{3}=2\left( \cos( 120+360n)+i \sin \left( (120+360n \right) \right)\] \[or Z ^{\frac{ 1 }{ 3 }}=2^1/3e ^{\frac{ 360n+120 }{ 3 }}\] put n=0,1,2 ,you get all the three cube roots
anonymous
  • anonymous
\[\left( 1+i \sqrt{3} \right)^{\frac{ 1 }{ 3 }}=2^{\frac{ 1 }{ 3 }}\left\{ \cos \left( \frac{ 360n+120 }{ 3 }+i \sin \left( \frac{ 360n+120 }{ 3 } \right)\right)\right\}\] put n=0,1,2 you get all the values.
anonymous
  • anonymous
Thank you so much! I very much appreciate it!
anonymous
  • anonymous
yw

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