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
"r^2=(\\sqrt{3})^2+(-1)^2=4""r\\ge0=>r=\\sqrt{4}=2"

Quadrant IV



"\\tan \\theta=\\dfrac{-1}{\\sqrt{3}}=-\\dfrac{1}{\\sqrt{3}}""\\theta=2\\pi+\\tan^{-1}(-\\dfrac{1}{\\sqrt{3}})=\\dfrac{11\\pi}{6}""z=2(\\cos\\dfrac{11\\pi}{6}+i\\sin\\dfrac{11\\pi}{6})"

"(2, \\dfrac{11\\pi}{6})"



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