Question 1135993
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The modulus  r = {{{(-sqrt(6))^2 + (sqrt(2))^2)}}} = {{{sqrt(6 + 2)}}} = {{{sqrt(8)}}} = {{{2*sqrt(2)}}}.


The argument  {{{tan(alpha)}}} = {{{-sqrt(2)/sqrt(6)}}} = {{{-1/sqrt(3)}}}  =============>  {{{alpha}}} = {{{11pi/6}}}.



Polar form:  {{{-sqrt(6) + sqrt(2)*i}}} = {{{(2*sqrt(2))*(cos(11pi/6) + i*sin(11pi/6))}}},   or   (r,{{{alpha}}}) = ({{{2*sqrt(2)}}},{{{11pi/6}}}).
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