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

Question

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

anonymous · Answer

@Ashleyisakitty @satellite73 @jim_thompson5910 @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate

anonymous · Answer

I'm worried theta might be wrong

zepdrix · Answer

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

anonymous · Answer

-2/-2=1 So I thought that was why. Is there a rule I don't know?

zepdrix · Answer

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 · Answer

Since they're both negative, we need the NEXT angle,
which will be pi/4 + pi

anonymous · Answer

Oh, I see what you mean.

anonymous · Answer

5pi/4?

anonymous · Answer

not to butt in, but drawing a picture is real helpful
by which i only mean "plot the point"

zepdrix · Answer

ya 5pi/4 sounds better :)

anonymous · Answer

Oh, gotcha @satellite73

anonymous · Answer

it will keep you from doing something silly like taking 
$$\tan^{-1}(\frac{b}{a})$$ which does not really work

anonymous · Answer

@satellite73 yup, I'll do that next time.

anonymous · Answer

Thanks for the help everyone :D