anonymous
  • anonymous
Extremely quick question! Thank You! Is w=-2-2i in polar form 2sqrt(2)(cos(pi/4)+isin(pi/4))?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Ashleyisakitty @satellite73 @jim_thompson5910 @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate
anonymous
  • anonymous
I'm worried theta might be wrong
zepdrix
  • zepdrix
hmm ya, it can't be pi/4 right? we're in quadrant `three` if both the real and imaginary part are negative.

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anonymous
  • anonymous
-2/-2=1 So I thought that was why. Is there a rule I don't know?
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
Since they're both negative, we need the NEXT angle, which will be pi/4 + pi
anonymous
  • anonymous
Oh, I see what you mean.
anonymous
  • anonymous
5pi/4?
anonymous
  • anonymous
not to butt in, but drawing a picture is real helpful by which i only mean "plot the point"
zepdrix
  • zepdrix
ya 5pi/4 sounds better :)
anonymous
  • anonymous
Oh, gotcha @satellite73
anonymous
  • anonymous
it will keep you from doing something silly like taking \[\tan^{-1}(\frac{b}{a})\] which does not really work
anonymous
  • anonymous
@satellite73 yup, I'll do that next time.
anonymous
  • anonymous
Thanks for the help everyone :D

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