Question

If $z=\frac{\sqrt{3}+i}{2}$, then $\left(z^{101}+i^{103}\right)^{105}$ is equal to

• A. $z$
• B. $z^{2}$
• C. $z^{3}$
• D. none of these

Answer 1

Answer: (C)

$ z=\frac{\sqrt{3}+i}{2}=\frac{i(1-\sqrt{3} i)}{2}=-i \omega$
$\therefore z^{101}+i^{103}=(-i \omega)^{101}+i^{103}=-i \omega^{2}-i=i \omega$
$\therefore \left(z^{101}+i^{103}\right)^{105}=(i \omega)^{105}=i$
Also, $z=-i \omega$
$ z^{2}=(-i \omega)^{2}=-\left(\omega^{2}\right)=-\omega^{2} \neq i$
$ z^{3}=(-i \omega)^{3}=-i^{3} \omega^{3}=i$
so, $\left(z^{101}+i^{103}\right)^{105}=z^{3}$