Find all the values of \(\left(\frac12+i\frac{\sqrt3}2\right)^{3/4}\) and show that their continued product is 1.

Step 1 (polar form): \(\frac12+i\frac{\sqrt3}{2}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\). Including the general angle, \(\frac12+i\frac{\sqrt3}{2}=\cos\left(2k\pi+\frac{\pi}{3}\right)+i\sin\left(2k\pi+\frac{\pi}{3}\right)\).

Step 2 (De Moivre): \(\left(\frac12+i\frac{\sqrt3}{2}\right)^{3/4}=\cos\frac{3(6k+1)\pi}{12}+i\sin\frac{3(6k+1)\pi}{12}=\cos\frac{(6k+1)\pi}{4}+i\sin\frac{(6k+1)\pi}{4},\quad k=0,1,2,3.\)

The four values:

\(k=0:\ \cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1+i}{\sqrt2}\)
\(k=1:\ \cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}=\frac{1-i}{\sqrt2}\)
\(k=2:\ \cos\frac{13\pi}{4}+i\sin\frac{13\pi}{4}=\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}=\frac{-1-i}{\sqrt2}\)
\(k=3:\ \cos\frac{19\pi}{4}+i\sin\frac{19\pi}{4}=\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}=\frac{-1+i}{\sqrt2}\)

(Check: \(\left(\frac12+i\frac{\sqrt3}{2}\right)^3=\cos\pi+i\sin\pi=-1\), and these are exactly the four fourth roots of \(-1\).)

Step 3 (continued product): Product \(=\cos\frac{(1+7+13+19)\pi}{4}+i\sin\frac{(1+7+13+19)\pi}{4}=\cos10\pi+i\sin10\pi=1\).

Continued product of the roots \(=1\).