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, $z=-i\omega$
$z^2=(-i\omega {)}^2=-\left({\omega }^2\right)=-{\omega }^2\ne i$
$z^3=(-i\omega {)}^3=-i^3{\omega }^3=i$
so, ${\left(z^{101}+i^{103}\right)}^{105}=z^3$