Question 1078163
{{{ ((-6i)/(-1+ sqrt( 3)i))( (-1-sqrt(3)i)/(-1- sqrt( 3)i)) =(6i+6sqrt(3)i^2)/(1-3i^2)}}}
{{{ ((-6i)/(-1+ sqrt( 3)i))( (-1-sqrt(3)i)/(-1- sqrt( 3)i)) =(-6sqrt(3)+6i)/(1+3)}}}
{{{ ((-6i)/(-1+ sqrt( 3)i))( (-1-sqrt(3)i)/(-1- sqrt( 3)i)) =6(-sqrt(3)+i)/(4)}}}
{{{ ((-6i)/(-1+ sqrt( 3)i))( (-1-sqrt(3)i)/(-1- sqrt( 3)i)) =-(3/2)(sqrt(3)-i)}}}
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So convert to polar coordinates,
(0,-6)
{{{R[1]=6}}}
{{{theta[1]=270}}}
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(-1,sqrt(3))
{{{R[2]=sqrt((-1)^2+(sqrt(3))^2)=sqrt(1+3)=sqrt(4)=2}}}
{{{theta[2]=tan^(-1)(sqrt(3)/1)=120}}}
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So then,
Magnitude:
{{{Z[1]/Z[2]=(R[1]/R[2])=6/2=3}}}
Angle:
{{{Z[1]/Z[2]=270-120=150}}}
All together,
{{{Z[1]/Z[2]=3cos(150)+i(3sin(150))}}}
{{{Z[1]/Z[2]=3(-sqrt(3)/2)+i(3(1/2))}}}
{{{Z[1]/Z[2]=(3/2)(-sqrt(3)+i)}}}
{{{Z[1]/Z[2]=-(3/2)(sqrt(3)-i)}}}