공학계열 3번

$\displaystyle \sqrt{3}+i=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)=2e^{i\frac{\pi}{6}}$

$\displaystyle (\sqrt{3}+i)^{12}=2^{12}e^{i\left(12\times\frac{\pi}{6}\right)}=2^{12}e^{i(2\pi)}$

$\displaystyle e^{i(2\pi)}=1$이므로 $\displaystyle 2^{12}e^{i(2\pi)}=2^{12}=4096$