Question 1138181
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(a) {{{(1-i*sqrt(3))(-1+i*sqrt(3)) = -1+i*sqrt(3)+i*sqrt(3)-3i^2 = 2+2i*sqrt(3)}}}<br>
(b)<br>
z1 = 1-1*sqrt(3):
{{{r = sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2}}}
{{{theta = arctan((-1*sqrt(3))/1) = -pi/3}}} (because z1 is in quadrant IV)
z1 = (2,-pi/3)<br>
z2 = -1+i*sqrt(3):
{{{r = sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2}}}
{{{theta = arctan((-1*sqrt(3))/1) = 2pi/3}}} (because z2 is in quadrant II)
z1 = (2,2pi/3)<br><br>
z1*z2 = (2,-pi/3)*(2,2pi/3) = (4,pi/3)<br>
c) (4,pi/3) = {{{4*cis(pi/3) = 4*(cos(pi/3)+i*sin(pi/3)) = 4*(1/2+i*sqrt(3)/2) = 2+2*sqrt(3)}}}