Express each of the following in the form (a + ib)

Solution:

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

Now, we rationalize the above equation by multiply and divide by the conjugate of (2 – i)

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

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

$=\frac{\left(4+9 i^{2}+12 i\right)(2+i)}{(2)^{2}-(i)^{2}}$

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

$=\frac{[4+9(-1)+12 i](2+i)}{4-i^{2}}\left[\because \cdot i^{2}=-1\right]$

$=\frac{[4-9+12 i](2+i)}{4-(-1)}$

$=\frac{(-5+12 i)(2+i)}{5}$

$=\frac{-10-5 i+24 i+12 i^{2}}{5}$

$=\frac{-10+19 i+12(-1)}{5}$

$=\frac{-10-12+19 i}{5}$

$=\frac{-22+19 i}{5}$

$=-\frac{22}{5}+\frac{19}{5} i$