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

Writing $\sqrt{3}+i$ in polar form and applying De Moivre's theorem reduces $(\sqrt3+i)^{100}$ to $2^{100}(\cos\theta+i\sin\theta)$ for some angle $\theta$; comparing with $2^{99}(a+ib)$ immediately gives $a^2+b^2=4$.

Step 1 — Polar form. $|\sqrt3+i| = \sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$, and $\arg(\sqrt3+i) = \tan^{-1}\left(\dfrac{1}{\sqrt3}\right) = \dfrac{\pi}{6}$.

So $\sqrt3+i = 2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$.

Step 2 — Apply De Moivre's theorem.

$(\sqrt3+i)^{100} = 2^{100}\left(\cos\dfrac{100\pi}{6}+i\sin\dfrac{100\pi}{6}\right) = 2^{100}(\cos\theta+i\sin\theta)$, where $\theta=\dfrac{50\pi}{3}$.

Step 3 — Compare with the given form.

We are told $(\sqrt3+i)^{100} = 2^{99}(a+ib)$. So …