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

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

Above equation can be re – written as

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

Now, rationalizing

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

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

$=\frac{4+i^{2}-4 i}{\left(4+i^{2}+4 i\right)\left(4+i^{2}-4 i\right)}\left[\because(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab}\right]$

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

$=\frac{3-4 i}{(3+4 i)(3-4 i)} \ldots(\mathrm{i})$

Now, we know that,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

So, eq. (i) become

$=\frac{3-4 i}{(3)^{2}-(4 i)^{2}}$

$=\frac{3-4 i}{9-16 i^{2}}$

$=\frac{3-4 i}{9-16(-1)}$

$=\frac{3-4 i}{25}$

$=\frac{3}{25}-\frac{4}{25} i$