Question 1172294
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If *[tex \Large z\ =\ a\ +\ bi], then *[tex \Large z\ =\ r\theta] where *[tex \Large r\ =\ \sqrt{a^2\,+\,b^2}] and *[tex \Large \theta\,=\,\arctan\(\frac{b}{a}\)].


For your problem, *[tex \Large a\ =\ -\sqrt{3}] and *[tex \Large b\ =\ -1].


Hint: *[tex \Large \arctan(x)\ =\ \tan^{-1}(x)], which is to say "the angle whose tangent is *[tex \Large x]"

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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