Answer to Question #350492 in Calculus for Nomsa

Question #350492

Convert (√3,-1)into polar coordinates (r,0) so that r≥0 and 0≤0<2π


1
Expert's answer
2022-06-14T12:24:37-0400
r2=(3)2+(1)2=4r^2=(\sqrt{3})^2+(-1)^2=4r0=>r=4=2r\ge0=>r=\sqrt{4}=2

Quadrant IV



tanθ=13=13\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{\sqrt{3}}θ=2π+tan1(13)=11π6\theta=2\pi+\tan^{-1}(-\dfrac{1}{\sqrt{3}})=\dfrac{11\pi}{6}z=2(cos11π6+isin11π6)z=2(\cos\dfrac{11\pi}{6}+i\sin\dfrac{11\pi}{6})

(2,11π6)(2, \dfrac{11\pi}{6})



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