Question 285171
i is defined as the {{{sqrt(-1)}}}
So we have.
{{{(1/3-sqrt(-1))^2}}} Now we FOIL.
{{{(1/3)*(1/3)-(1/3)sqrt(-1)-(1/3)sqrt(-1)+sqrt(-1)*sqrt(-1)}}}
Simplify.
{{{1/9-(2/3)sqrt(-1)+(-1)}}} note that {{{sqrt(-1)*sqrt(-1)=-1}}}
And now put {{{sqrt(-1)}}} back as i since that is what teachers expect.
{{{1/9-(2/3)i-1}}}
So {{{(1/3 - i)^2=1/9-(2/3)i-1}}}