Question

If $(\sqrt{3}+i{)}^{100}=2^{99}(a+ib)$, then show that $a^2+b^2=4$

Answer 1

Since $\omega =\frac{-1+i\sqrt{3}}{3}=\frac{i^2+i\sqrt{3}}{2}$

$\Rightarrow \sqrt{3}+i=\frac{2\omega }{i}=-2\omega i$

Thus $(\sqrt{3}+i{)}^{100}=(-2\omega i{)}^{100}=2^{100}\cdot {\omega }^{100}{i}^{100}$

$=2^{100}\omega \left(1\right)=2^{100}\left(\frac{-1+i\sqrt{3}}{2}\right)=2^{99}(-1+i\sqrt{3})$

So $a=-1$ and $b=\sqrt{3}$

Hence $a^2+b^2=1+3=4$