write z=1 - √3 i in polar form I got my r = 2 and i got arg (theta) value as -pi/3

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write z=1 - √3 i in polar form I got my r = 2 and i got arg (theta) value as -pi/3

Mathematics
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the answer reads: (g) The modulus of 1 - √3 i is √12 + (√3)2 = 2 and its argument θ satisfies tan θ = - √3/1 = - √3, whence θ is equal to either 2π/3 or 5π/3. Since y=-√3 is negative, we have θ = 5π/3. The polar form of -3 + 3i is thus 2 (cos 5π/3 + i sin 5π/3). ....
|dw:1441322582245:dw| Thats what i thought
this example too seems to be adding when i thought arg (theta) is positive \[-\pi < \theta \le \pi \]
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Other answers:

unless im just confused.
im still confused
that has to be the ugliest circle i have ever seen
correct answer though
Umm no that's not a circle
I was trying to show that when the Arg theta value is between 0 and 180deg then it's positive
|dw:1441326450077:dw| your point is in the 4th quadrant
|dw:1441326527937:dw|
Yes so should it be a negative and I got negative pi/3
|dw:1441326541656:dw|
right -pi/3 is right
but
Why haa the model answera added 3pi/2 to pi/3
it looks like they wanted you to write theta in between 0 and 2pi based on their answer
so you just do -pi/3+2pi
this will put you between 0 and 2pi
i see you wanted to add pi for some reason but that would put you in the second quadrant
|dw:1441326754891:dw| doing theta+pi will put you in second quadrant as you can see
and we definitely want to be in 4th
theta+2pi would get us back there to fourth quadrant
If the question doesn't specify will -pi/3 also be a sufficient answer
yes if they don't specify what they want theta between yes -pi/3 will do and so would -pi/3+2pi*n where n is an integer as long as you keep your r positive that is
Also can u look at the image I have attached above...I guess they have demonstrated both methods
\[(r,\theta)=(r,\theta+2 \pi n) \\ \text{ or } \\ (r,\theta)=(-r,\theta+\pi+2 \pi n)\]
when it comes to polar coordinates there is not a unique way to express one point
not a unique way meaning there are infinite amount of ways to express one point
so you could haven chosen (-2,2pi/3)
on account of sine and cosine are periodic functions with period \(2\pi\)
or (2,5pi/3) or (2,-pi/3)
and so on

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