Question

Q.4. Write the multiplicative inverse of $\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.i$

Answer 1

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} \frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i} \\ =\frac{\sqrt{3}{\displaystyle -}{\displaystyle \mathrm{i}}}{2} \\ \mathrm{And}\mathrm{multiplicative}\mathrm{inverse}\mathrm{of}\frac{{\displaystyle \sqrt{3}-\mathrm{i}}}{2}\mathrm{is}\frac{2}{\sqrt{3}{\displaystyle -}{\displaystyle \mathrm{i}}}. \\ \mathrm{And},\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}-\mathrm{i}}}=\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}-\mathrm{i}}}\times \frac{\sqrt{3}{\displaystyle +}{\displaystyle \mathrm{i}}}{\sqrt{3}{\displaystyle +}{\displaystyle \mathrm{i}}}\left\{\mathrm{rationalising}\right\} \\ =\frac{2\left(\sqrt{3}+\mathrm{i}\right)}{{\displaystyle {\left(\sqrt{3}\right)}^2}{\displaystyle -}{\displaystyle {{\displaystyle \mathrm{i}}}^2}} \\ =\frac{2\sqrt{3}+2\mathrm{i}}{3-\left(-1\right)}\{\mathrm{since},{\mathrm{i}}^2=-1\} \\ =\frac{2\sqrt{3}+2\mathrm{i}}{4} \\ =\frac{2\sqrt{3}}{4}+\frac{2\mathrm{i}}{4} \\ =\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2} \\ =\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 2}}\mathrm{i} \\ \mathrm{Hence}\mathrm{multiplicative}\mathrm{inverse}\mathrm{of}\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}\mathrm{i}\mathrm{is}\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 2}}\mathrm{i}. \end{gathered}$$
Regards