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

- anonymous

- jamiebookeater

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- anonymous

I think it's false

- freckles

|dw:1438653320969:dw|
let's think about rewriting this polar coordinate so that theta is between 0 and 360

- freckles

|dw:1438653367482:dw|

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## More answers

- freckles

but what happens when you change the sign of that other r

- freckles

for example where is (-3,225 deg)

- anonymous

\[(3\sqrt{2}/2, 3\sqrt{2}/2)\] ?

- anonymous

wait it would be -3root2/2, -3root2/2 right?

- freckles

|dw: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

- anonymous

Oh okay yes

- freckles

if 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

as you they both converted to the same Cartesian coordinate pair

- freckles

and both have their theta's between 0 and 360 deg

- anonymous

So they are not all unique

- freckles

When 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

If 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

Okay so it would be true

- anonymous

Awesome! Thank you so much

- freckles

right not unique
unique means we can only find one but we found two

- anonymous

hmmm @freckles it was false. Should I ask my teacher if the question is wrong?

- freckles

it 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

yep 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

Ya 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

np

- freckles

http://openstudy.com/study#/updates/53796108e4b0e5430795a83f
this question is almost the same question

- freckles

i 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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