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tan^-1(-sqrt 3)
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arctan(-sqrt(3)) = x <==> tan(x) = -sqrt(3) give me an angle whose tangent is -sqrt(3)
Im pretty sure that you can just plug that into a calculator
z=-1+√3i Θ=ArgZ=tan^-1(-√3 )=-pi/3 ArgZ=?
?????
Complex number : Z=re^iArgZ z=-1+√3i , r=|Z|=√a^2+b^2=√4=2 Θ=ArgZ=tan^-1(-√3 )=-pi/3 ArgZ=?
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is this even required? It's just a simple trig question
Which area of the X and Y axis is
well, -pi/3 is correct :DDD
yes, tank u
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