Question

Express the mentioned below complex number in the standard form $a+ib$
(viii) $\frac{(1-i{)}^3}{1-i^3}$

Answer 1

$\because \frac{(1-i{)}^3}{1-i^3}$

$=\frac{1^3-i^3-3\times 1^2\times i+3\times 1\times i^2}{1-(-i)}$

$=\frac{1-(-i)-3i+3i^2}{1+i}$

$=\frac{-2-2i}{1+i}$

$=\frac{-2(1+i)}{1+i}$

$=-2+0.i$