Question 74789
{{{12cos(pi/3)+i*12sin(pi/3))}}}Distribute the 12
{{{12cos(pi/3)+i*12sin(pi/3))}}}Evaluate {{{cos(pi/3)}}}and {{{sin(pi/3)}}}
{{{12(1/2)+i*12*(sqrt(3)/2)}}}
{{{6+i*6*(sqrt(3))}}}
So {{{12cos(pi/3)+i*12sin(pi/3))}}} becomes
{{{6+6*sqrt(3)i}}}
Which is in {{{a+bi}}} form where {{{a=6}}} and {{{b=6sqrt(3)}}}