Question

write z=1 - √3 i in polar form

I got my r = 2
and i got arg (theta) value as -pi/3

Answer 1

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).


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Answer 2

Thats what i thought