George C. answered • 12/02/12

Humboldt State and Georgetown graduate

r =(((-4)(3)^(1/2))^2 + ((-4)^2))^(1/2) = 8

θ = arctan (-4/(-4(3)^(1/2) = 7∏/6.

roots = 2 (cos (7∏/18 + 12n∏/18) = i sin(7∏/18 + 12n∏/18), n = 0, 1, 2

Diana L.

asked • 12/02/12I have to remember to convert -4√3 – 4i) in to polar form

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George C. answered • 12/02/12

Humboldt State and Georgetown graduate

r =(((-4)(3)^(1/2))^2 + ((-4)^2))^(1/2) = 8

θ = arctan (-4/(-4(3)^(1/2) = 7∏/6.

roots = 2 (cos (7∏/18 + 12n∏/18) = i sin(7∏/18 + 12n∏/18), n = 0, 1, 2

Roman C. answered • 12/02/12

Masters of Education Graduate with Mathematics Expertise

z = 8 cis (2πk-5π/6) for k = 1,2,3.

z^{1/3} = 2 cis (2πk/3 - 5π/18) for k = 1,2,3.

= 2 cos(π/18) + 2i sin(π/18)

or 2 cos(7π/18) + 2i sin(7π/18)

or 2 cos(13π/18) + 2i sin(13π/18)

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George C.

roots = 2 (cos (7?/18 + 12n?/18) + i sin(7?/18 + 12n?/18), n = 0, 1, 2

I didn't hit <Shift> when pressing the (+) sign and got (=). Right button, wrong function.

12/03/12