Solve for me those Questions?

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  • 1 month ago

    (a)  r = 1, theta = 2pi/9.

    (b)  First multiply top and bottom by +sqrt(3) + i, you get[2*sqrt(3) + 2i*sqrt(3) + 2i - 2]/(-3 - 1)= -sqrt(3)/2 + 1/2 + i[-sqrt(3)/2 - 1/2].r = sqrt{[3/4 + 1/4 - sqrt(3)/2] + [3/4 + 1/4 + sqrt(3)/2]}= sqrt(2).theta = arctan[(-sqrt(3)/2 - 1/2)/(-sqrt(3)/2 + 1/2)]= about 258 degrees, I don't know what form they want.(c)  4i/(9e^(4+i))= 4e^(i*pi/2)/(9e^(4+i)).r = 4/(9e^4).Then you have e^(i*theta) = e^(i*pi/2)/e^i= e^[i(pi/2 - 1)],so theta = pi/2 - 1.

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