## 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.

1. anonymous

I think it's false

2. freckles

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

3. freckles

|dw:1438653367482:dw|

4. freckles

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

5. freckles

for example where is (-3,225 deg)

6. anonymous

$(3\sqrt{2}/2, 3\sqrt{2}/2)$ ?

7. anonymous

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

8. 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

9. anonymous

Oh okay yes

10. freckles

if you want to write both and Cartesian coordinates you can... $(r=3,\theta=45 \deg) \\ x=r cos(\theta)= 3 \cos(45) =3 \frac{\sqrt{2}}{2} =\frac{ 3 \sqrt{2}}{2} \\ y =r \sin(\theta)= 3 \sin(45) =3 \frac{\sqrt{2}}{2}= \frac{3 \sqrt{2}}{2} \\ (r=-3, \theta=225 \deg) \\ x =rcos(\theta)=-3 \cos(225)=-3 \frac{-\sqrt{2}}{2}=\frac{3 \sqrt{2}}{2} \\ y=r \sin(\theta)= -3 \sin(225) -3 \frac{-\sqrt{2}}{2}=\frac{3 \sqrt{2}}{2}$

11. freckles

as you they both converted to the same Cartesian coordinate pair

12. freckles

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

13. anonymous

So they are not all unique

14. 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.

15. 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

16. anonymous

Okay so it would be true

17. anonymous

Awesome! Thank you so much

18. freckles

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

19. anonymous

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

20. freckles

it is true since $(r,\theta)=(-r, \theta+\pi) \text{ if } \theta \in [0,\pi) \\ (r,\theta)=(-r,\theta-\pi) \text{ if } \theta \in [\pi,2\pi]$

21. 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

22. 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!

23. freckles

np

24. freckles

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

25. 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