Find the conjugate of each of the following:

Given: $\frac{(1+i)(2+i)}{(3+i)}$

Firstly, we calculate $\frac{(1+i)(2+i)}{(3+i)}$ and then find its conjugate

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

$=\frac{2+i+2 i+i^{2}}{3+i}$

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

$=\frac{1+3 i}{3+i}$

Now, we rationalize the above by multiplying and divide by the conjugate of 3 + i

$=\frac{1+3 i}{3+i} \times \frac{3-i}{3-i}$

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

Now, we know that,

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

So, eq. (i) become

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

$=\frac{1(3)+1(-i)+3 i(3)+3 i(-i)}{9-i^{2}}$

$=\frac{3-i+9 i-3 i^{2}}{9-(-1)}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{3+8 i-3(-1)}{9+1}\left[\because j^{2}=-1\right]$

$=\frac{3+8 i+3}{10}$

$=\frac{6+8 i}{10}$

$=\frac{2(3+4 i)}{10}$

$=\frac{3+4 i}{5}$

$=\frac{3}{5}+\frac{4}{5} i$

Hence, $\frac{(1+i)(2+i)}{(3+i)}=\frac{3}{5}+\frac{4}{5} i$

So, the conjugate of $\frac{(1+i)^{2}}{(3-i)}$ is $\frac{3}{5}-\frac{4}{5} i$