Find the modulus of each of the following:

Solution:

Given: $\frac{(3+2 i)^{2}}{(4-3 i)}$

Firstly, we calculate $\frac{(3+2 i)^{2}}{(4-3 i)}$ and then find its modulus

$\frac{(3+2 i)^{2}}{(4-3 i)}=\frac{9+4 i^{2}+12 i}{(4-3 i)}\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\frac{9+4(-1)+12 i}{4-3 i}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{5+12 i}{4-3 i}$

Now, we rationalize the above by multiplying and divide by the conjugate of 4 + 3i

$=\frac{5+12 i}{4-3 i} \times \frac{4+3 i}{4+3 i}$

$=\frac{(5+12 i)(4+3 i)}{(4-3 i)(4+3 i)} \ldots(\mathrm{i})$

Now, we know that,

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

So, eq. (i) become

$=\frac{5(4)+(5)(3 i)+12 i(4)+12 i(3 i)}{(4)^{2}-(3 i)^{2}}$

$=\frac{20+15 i+48 i+36 i^{2}}{16-9 i^{2}}$

$=\frac{20+63 i+36(-1)}{16-9(-1)}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{20-36+63 i}{16+9}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{-16+63 i}{25}$

$=-\frac{16}{25}+\frac{63}{25} i$

Now, we have to find the modulus of $\left(-\frac{16}{25}+\frac{63}{25} i\right)$

So, $|z|=\left|-\frac{16}{25}+\frac{63}{25} i\right|=\sqrt{\left(-\frac{16}{25}\right)^{2}+\left(\frac{63}{25}\right)^{2}}$

$=\sqrt{\frac{256}{625}+\frac{3969}{625}}$

$=\sqrt{\frac{4225}{625}}$

$=\frac{65}{25}$

$=\frac{13}{5}$

Hence, the modulus of $\frac{(3+2 i)^{2}}{(4-3 i)}$ is $\frac{13}{5}$