Find the conjugate of each of the following:

Solution:

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

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

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

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

$=\frac{2 i}{3-i}$

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

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

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

Now, we know that,

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

So, eq. (i) become

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

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

$=\frac{6 i+2 i^{2}}{9-(-1)}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{6 i+2(-1)}{9+1}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{6 i-2}{10}$

$=\frac{2(3 i-1)}{10}$

$=\frac{(-1+3 i)}{5}$

$=-\frac{1}{5}+\frac{3}{5} i$

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

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