Question 117942
{{{(sqrt(3)+i)^9}}} Start with the given expression



{{{(2(sqrt(3)/2+i(1/2)))^9}}}  Replace {{{sqrt(3)}}} and {{{i}}} with {{{sqrt(3)/2}}} and {{{i(1/2)}}}. Then place a 2 outside the parenthesis. Notice how the 2 on the outside will distribute back to {{{sqrt(3)+i}}}. In other words, {{{2(sqrt(3)/2+i(1/2))=sqrt(3)+i}}}



Notice how {{{cos(pi/6)=sqrt(3)/2}}} and {{{sin(pi/6)=1/2}}}.


{{{(2(cos(pi/6)+i*sin(pi/6)))^9}}} Now replace {{{sqrt(3)/2}}} with {{{cos(pi/6)}}} and replace {{{1/2}}} with {{{sin(pi/6)}}}.


Now use De Moivre's Thereom to expand. Remember, De Moivre's Thereom states:


{{{(r(cos(x)+i*sin(x)))^n=r^n*(cos(nx)+i*sin(nx))}}}



{{{2^9(cos(9(pi/6))+i*sin(9(pi/6)))}}} Expand using De Moivre's Thereom



{{{512(cos(3pi/2)+i*sin(3pi/2))}}} Simplify



{{{512(0+i*(-1))}}} Now evaluate {{{cos(3pi/2)}}} and {{{sin(3pi/2)}}}



{{{0-512i}}} Distribute



So the answer is now in {{{a+bi}}} form where {{{a=0}}} and {{{b=-512}}}



So {{{(sqrt(3)+i)^9=0-512i}}}