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convert -sqrt(3)-i into polar form
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\[ z= x+iy \implies x = -\sqrt 3,\quad y = -1 \]So use: \[ z=re^{i\theta } \implies r = \sqrt{x^2+y^2},\quad \theta = \tan^{-1}\left(\frac{-1}{ -\sqrt 3}\right) \]
2cis (pi)/6 2cis 5(pi) 6 2cis 7 6 2cis 11 6 2cis 4 3
r=-sqrt(3)^2+-1^2
im sorry i'm exhausted haven't been to sleep in 2 days man I'm about to be out
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