Find the multiplicative inverse of each of the following:

Solution:

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

To find: Multiplicative inverse

We know that

Multiplicative Inverse of $z=z^{-1}$

$=\frac{1}{z}$

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

So, Multiplicative inverse of $\frac{(1+i)(1+2 i)}{(1+3 i)}=\frac{1}{\frac{(1+i)(1+2 i)}{(1+3 i)}}$

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

We solve the above equation

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

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

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

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

Now, we rationalize the above by multiplying and divide by the conjugate of (-1 + 3i)

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

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

Now, we know that,

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

So, eq. (i) become

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

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

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

$=\frac{-1-6 i+9}{1+9}$

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

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

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

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

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