If $\mathrm{z}=\frac{1}{2}(\sqrt{3}-\mathrm{i})$ and the smallest value of the positive integer n for which ${\left({\mathrm{z}}^{89}+{\mathrm{i}}^{97}\right)}^{94}={\mathrm{z}}^{\mathrm{n}}$ is $\mathrm{\lambda }$ then $\frac{\mathrm{\lambda }}{2}$ = 

We have $\mathrm{z}=\frac{1}{2}(\sqrt{3}-\mathrm{i})=-\frac{1}{2}\mathrm{i}(1+\mathrm{i}\sqrt{3})={\mathrm{i\omega }}^2$ where $\mathrm{\omega }\ne 1$ is a cube root of unity.
$\therefore {\mathrm{z}}^{89}={\left({\mathrm{i\omega }}^2\right)}^{89}={\mathrm{i}}^{89}{\mathrm{\omega }}^{178}=\mathrm{i\omega }\text{ Also, }{\mathrm{i}}^{97}={\mathrm{i}}^{96}\mathrm{i}=\mathrm{i}$
Thus, ${\left({\mathrm{z}}^{89}+{\mathrm{i}}^{97}\right)}^{94}=(\mathrm{i\omega }+\mathrm{i}{)}^{94}={\left[\mathrm{i}\left(-{\mathrm{\omega }}^2\right)\right]}^{94}=-{\mathrm{\omega }}^2$
$\therefore$ The given equation becomes
$-{\mathrm{\omega }}^2={\mathrm{i}}^{\mathrm{n}}{\mathrm{\omega }}^{2\mathrm{n}}\Rightarrow {\mathrm{i}}^{\mathrm{n}}{\mathrm{\omega }}^{2\mathrm{n}-2}=-1$
Therefore, $\mathrm{n}=10\text{ and }\mathrm{\lambda }=10\Rightarrow \frac{\mathrm{\lambda }}{2}=5\text{. }$