Question

Solve it: ${\left(1+i\right)}^{-3}$

Answer 1

${\left(1+i\right)}^{-3}\Rightarrow \frac{1}{{\left(1+i\right)}^3}$

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

$\frac{-1}{2-2i}$(multiply $\left(2+2i\right)$ in numerator and denominator both)

$\frac{-1\times \left(2+2i\right)}{\left(2-2i\right)(2+2i)}=\frac{-2-2i}{2^2-{\left(2i\right)}^2}=\frac{-2-2i}{4-4i^2}=\frac{-2\left(1+i\right)}{4+4}=\frac{-2\left(1+i\right)}{8}=-\frac{1}{4}\left(1+i\right)$

Hence the answer is $-\frac{1}{4}\left(1+i\right)$