Question

Express the given complex number in the form $a+ib:$${(\frac{1}{3}+3i)}^3$

Answer 1

Using $(a+b{)}^3=a^3+b^3+3ab(a+b)$ we have,
${(\frac{1}{3}+3i)}^3={\left(\frac{1}{3}\right)}^3+{\left(3i\right)}^3+3\left(\frac{1}{3}\right)\left(3i\right)(\frac{1}{3}+3i)$
$=\frac{1}{27}+27i^3+3i(\frac{1}{3}+3i)$
$=\frac{1}{27}+27(-i)+i+9i^2,[\because i^3=-i]$
$=\frac{1}{27}-27i+i-9,[\because i^2=-1]$
$=(\frac{1}{27}-9)+i(-27+1)$
$=\frac{-242}{27}-26i=a+ib$
where $a=\frac{-242}{27}$ and $b=-26$