Find the multiplicative inverse of each of the following:

Solution:

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

To find: Multiplicative inverse

We know that

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

$=\frac{1}{z}$

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

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

Now, rationalizing by multiply and divide by the conjugate of (2+3i)

$=\frac{1+i}{2+3 i} \times \frac{2-3 i}{2-3 i}$

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

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

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

$=\frac{2-3 i+2 i-3 i^{2}}{4-9 i^{2}}$

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

$=\frac{5-i}{4+9}$

$=\frac{5-i}{13}$

$=\frac{5}{13}-\frac{1}{13} i$

Hence, Multiplicative Inverse of $\frac{(2+3 i)}{1+i}$ is $\frac{5}{13}-\frac{1}{13} i$