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anonymous

  • one year ago

Extremely quick question! Thank You! Is w=-2-2i in polar form 2sqrt(2)(cos(pi/4)+isin(pi/4))?

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  1. anonymous
    • one year ago
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    @Ashleyisakitty @satellite73 @jim_thompson5910 @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate

  2. anonymous
    • one year ago
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    I'm worried theta might be wrong

  3. zepdrix
    • one year ago
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    hmm ya, it can't be pi/4 right? we're in quadrant `three` if both the real and imaginary part are negative.

  4. anonymous
    • one year ago
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    -2/-2=1 So I thought that was why. Is there a rule I don't know?

  5. zepdrix
    • one year ago
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    Well, when you apply your arctangent (im assuming that's what you're doing),\[\Large\rm \arctan\left(\frac{-2}{-2}\right)=\theta\]Yes, you end up with pi/4 as a reference angle. But this really corresponds to two angles in one full rotation. You have to look back at your rectangular form to see where you should be.

  6. zepdrix
    • one year ago
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    Since they're both negative, we need the NEXT angle, which will be pi/4 + pi

  7. anonymous
    • one year ago
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    Oh, I see what you mean.

  8. anonymous
    • one year ago
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    5pi/4?

  9. anonymous
    • one year ago
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    not to butt in, but drawing a picture is real helpful by which i only mean "plot the point"

  10. zepdrix
    • one year ago
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    ya 5pi/4 sounds better :)

  11. anonymous
    • one year ago
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    Oh, gotcha @satellite73

  12. anonymous
    • one year ago
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    it will keep you from doing something silly like taking \[\tan^{-1}(\frac{b}{a})\] which does not really work

  13. anonymous
    • one year ago
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    @satellite73 yup, I'll do that next time.

  14. anonymous
    • one year ago
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    Thanks for the help everyone :D

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