Simplify each of the following and express it in the form (a + ib) :

Solution:

Given: $(1+2 i)^{-3}$

Above equation can be re - written as

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

Now, rationalizing

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

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

We know that,

$(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$

$(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$

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

$=\frac{1-6 i+6 i^{2}-8 i^{3}}{\left[1+6 i+6 i^{2}+8 i^{3}\right]\left[1-6 i+6 i^{2}-8 i^{3}\right]}$

$=\frac{1-6 i+6(-1)-8 i(-1)}{[1+6 i+6(-1)+8 i(-1)][1-6 i+6(-1)-8 i(-1)]}\left[\because i^{2}=-1\right]$

$=\frac{1-6 i-6+8 i}{[1+6 i-6-8 i][1-6 i-6+8 i]}$

$=\frac{-5+2 i}{[-5-2 i][-5+2 i]}$

$=\frac{-5+2 i}{-5(-5)-5(2 i)-2 i(-5)-2 i(2 i)}$

$=\frac{-5+2 i}{25-10 i+10 i-4 i^{2}}$

$=\frac{-5+2 i}{25-4(-1)}\left[\because i^{2}=-1\right]$

$=\frac{-5+2 i}{29}$

$=-\frac{5}{29}+\frac{2}{29} i$