Solution:
(i) 
Given that
It is known that the multiplicative inverse of a complex number 
On multiplying and dividing by 
![\begin{array}{l} =\frac{1}{1-i} \times \frac{1+i}{1+i} \\ =\frac{1+i}{1^{2}-(i)^{2}} \\ =\frac{1+i}{1-(-1)}\left[\text { Since, } \mathrm{i}^{2}=-1\right] \\ =\frac{1+i}{2} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-50fbcdc7b79af3b8a71aaf14d8c3581a_l3.png "Rendered by QuickLaTeX.com")
As a result, the multiplicative inverse of 
(ii) 
Given that
![\begin{array}{l} (1+i \sqrt{3})^{2} \\ Z=(1+i \sqrt{3})^{2} \\ =1^{2}+(i \sqrt{3})^{2}+2(1)(i \sqrt{3}) \\ =1+3 i^{2}+2 i \sqrt{3} \\ =1+3(-1)+2 i \sqrt{3}\left[\text { since, } i^{2}=-1\right] \\ =1-3+2 i \sqrt{3} \\ =-2+2 i \sqrt{3} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5c55c8a67f7e56085397d5a67e9543e0_l3.png "Rendered by QuickLaTeX.com")
It is known that the multiplicative inverse of a complex number
Therefore,
On multiplying and dividing by 
As a result, the multiplicative inverse of 
