$\text{ If }Z=\frac{\sqrt{3}+i}{2},\text{ then }{\left[Z^{101}+i^{103}\right]}^{105}\text{ is equal to }$

$\begin{array}{r}\text{ Given, }z=\frac{\sqrt{3}+i}{2}=\frac{i[1-i\sqrt{3}]}{2}=-iw\\ \text{ Consider, }{z}^{101}+i^{103}=(-iw{)}^{101}+i^{103}=-i{w}^2-i=iw\\ \text{ Now, }{\left[z^{101}+i^{103}\right]}^{105}=(iw{)}^{105}=i\\ \text{ Also }z=-iw\\ z^2=(-iw{)}^2=i^2{w}^2=-w^2\\ z^3=(-iw{)}^3=-i^3\cdot w^3=i\\ \left[z^{101}+i^{103}\right]=z^3\end{array}$