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

Converting to polar form and using De Moivre gives $a=-1,\ b=\sqrt3$, so $a^2+b^2=4$.

$\sqrt3+i$ has modulus $r=\sqrt{3+1}=2$ and amplitude $\theta$ with $\cos\theta=\tfrac{\sqrt3}{2},\ \sin\theta=\tfrac12$, i.e. $\theta=30^\circ$.

So $\sqrt3+i = 2(\cos30^\circ+i\sin30^\circ)$.

By De Moivre's theorem, $(\sqrt3+i)^{100}=2^{100}(\cos3000^\circ+i\sin3000^\circ)$.

Since $3000^\circ = 8(360^\circ)+120^\circ$, the angle reduces to $120^\circ$: …