Find all the values of \(\left(\frac12+i\frac{\sqrt3}{2}\right)^{3/4}\) and show that their continued product is 1. (asked in MU Dec 2025; also in June 2025 KT)

Answer: The four values are \(\pm\frac{1}{\sqrt2}\pm\frac{i}{\sqrt2}\), i.e. \(e^{i\pi/4},e^{i3\pi/4},e^{i5\pi/4},e^{i7\pi/4}\), and their product is 1.

\(\frac12+i\frac{\sqrt3}{2}=\cos\frac\pi3+i\sin\frac\pi3\).

\(\left(\cos\frac\pi3+i\sin\frac\pi3\right)^{3}=\cos\pi+i\sin\pi\) (De Moivre). So we need the fourth roots of \(\cos\pi+i\sin\pi\):

\[\left(\tfrac12+i\tfrac{\sqrt3}{2}\right)^{3/4}=\left[\cos(2k\pi+\pi)+i\sin(2k\pi+\pi)\right]^{1/4}=\cos\frac{(2k+1)\pi}{4}+i\sin\frac{(2k+1)\pi}{4},\ k=0,1,2,3.\]

\(k=0:\ \frac1{\sqrt2}+\frac{i}{\sqrt2}\); \(k=1:\ -\frac1{\sqrt2}+\frac{i}{\sqrt2}\); \(k=2:\ -\frac1{\sqrt2}-\frac{i}{\sqrt2}\); \(k=3:\ \frac1{\sqrt2}-\frac{i}{\sqrt2}\).

Continued product: multiplying the four cis-forms adds the arguments:

\[\text{Product}=\cos\left(\tfrac{\pi+3\pi+5\pi+7\pi}{4}\right)+i\sin\left(\tfrac{16\pi}{4}\right)=\cos4\pi+i\sin4\pi=1.\]

Hence the continued product of all the values is 1.