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anonymous
 one year ago
True or False: When its argument is restricted to [0, 2pi), the polar form of a complex number is not unique.
anonymous
 one year ago
True or False: When its argument is restricted to [0, 2pi), the polar form of a complex number is not unique.

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438653320969:dw let's think about rewriting this polar coordinate so that theta is between 0 and 360

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438653367482:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but what happens when you change the sign of that other r

freckles
 one year ago
Best ResponseYou've already chosen the best response.2for example where is (3,225 deg)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(3\sqrt{2}/2, 3\sqrt{2}/2)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait it would be 3root2/2, 3root2/2 right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438653913563:dw I was going for that if you change the sign of the r it goes in the opposite direction on that same line there

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if you want to write both and Cartesian coordinates you can... \[(r=3,\theta=45 \deg) \\ x=r cos(\theta)= 3 \cos(45) =3 \frac{\sqrt{2}}{2} =\frac{ 3 \sqrt{2}}{2} \\ y =r \sin(\theta)= 3 \sin(45) =3 \frac{\sqrt{2}}{2}= \frac{3 \sqrt{2}}{2} \\ (r=3, \theta=225 \deg) \\ x =rcos(\theta)=3 \cos(225)=3 \frac{\sqrt{2}}{2}=\frac{3 \sqrt{2}}{2} \\ y=r \sin(\theta)= 3 \sin(225) 3 \frac{\sqrt{2}}{2}=\frac{3 \sqrt{2}}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2as you they both converted to the same Cartesian coordinate pair

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and both have their theta's between 0 and 360 deg

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So they are not all unique

freckles
 one year ago
Best ResponseYou've already chosen the best response.2When its argument is restricted to [0,2pi), we can still find 2 polar points one with a negative r and one with a positive r. So not unique. If they had said: When its argument is restricted to [0,2pi) where r>0, then the polar coordinate is unique.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2If they had said: When its argument is restricted to [0,2pi) where r>0, then the polar coordinate is unique. then this would be true <forgot to add this last line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so it would be true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! Thank you so much

freckles
 one year ago
Best ResponseYou've already chosen the best response.2right not unique unique means we can only find one but we found two

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm @freckles it was false. Should I ask my teacher if the question is wrong?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it is true since \[(r,\theta)=(r, \theta+\pi) \text{ if } \theta \in [0,\pi) \\ (r,\theta)=(r,\theta\pi) \text{ if } \theta \in [\pi,2\pi]\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep in either case there is no one way to write the complex number with restriction that theta has to be between 0 and 2pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya I understand the explanation. Should be right. So I'll just have to ask her next time I see her! Thank you for your help!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://openstudy.com/study#/updates/53796108e4b0e5430795a83f this question is almost the same question

freckles
 one year ago
Best ResponseYou've already chosen the best response.2i wonder if maybe they were just changing things on the homework to make it not identical but maybe forgot to change some of the answers
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