Answer to Question #122815 in Trigonometry for Ojugbele Daniel

Question

Simplify (cos2π/3 + isin2π/3)^1/4 and express the result in a form free from trigonometrically expression.

Answer 1

We will use the formula for roots of the complex numbers (https://en.wikipedia.org/wiki/De_Moivre%27s_formula#Roots_of_complex_numbers).

For $z=r(cos\,x+i\,sin\,x), r>0$ , n-th root is given by:

$z^{\frac{1}{n}}=r^{\frac1n}(cos\,\frac{x+2\pi k}{n}+i\,sin\,\frac{x+2\pi k}{n}),\,\,k=0,...,n-1$

Thus, for n=4, and $z=cos\,\frac{2\pi}{3}+i\,sin\,\frac{2\pi}{3}$ we receive:

$z^{\frac{1}{4}}=cos\,\frac{\frac{2\pi}{3}+2\pi k}{4}+i\,sin\,\frac{\frac{2\pi}{3}+2\pi k}{4}= cos\,(\frac{\pi}{6}+\frac{\pi k}{2})+i\,sin\,(\frac{\pi}{6}+\frac{\pi k}{2}),\,\,k=0,...,3$

The latter provides the following values:

$(z^{\frac{1}{4}})_0=cos\,\frac{\pi}{6}+i\,sin\,\,\frac{\pi}{6}=\frac{\sqrt{3}}{2}+i\,\frac{1}{2}\approx0.87\,+i\,\,\,0.5$

$(z^{\frac{1}{4}})_1=cos\,\frac{2\pi}{3}+i\,sin\,\,\frac{2\pi}{3}=-\frac{{1}}{2}+i\,\frac{\sqrt{3}}{2}\approx-0.5+i\,\,\,0.87$

$(z^{\frac{1}{4}})_2=cos\,\frac{7\pi}{6}+i\,sin\,\,\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}-i\,\frac{1}{2}\approx-0.87\,-i\,\,\,0.5$

$(z^{\frac{1}{4}})_3=cos\,\frac{5\pi}{3}+i\,sin\,\,\frac{5\pi}{3}=\frac{{1}}{2}-i\,\frac{\sqrt{3}}{2}\approx0.5-i\,\,\,0.87$

Answer: $(z^{\frac{1}{4}})_0=\frac{\sqrt{3}}{2}+i\,\frac{1}{2}\approx0.87\,+i\,\,\,0.5$ ; $(z^{\frac{1}{4}})_1= -\frac{{1}}{2}+i\,\frac{\sqrt{3}}{2}\approx-0.5+i\,\,\,0.87$ ; $(z^{\frac{1}{4}})_2= -\frac{\sqrt{3}}{2}-i\,\frac{1}{2}\approx-0.87\,-i\,\,\,0.5$ ; $(z^{\frac{1}{4}})_3= \frac{{1}}{2}-i\,\frac{\sqrt{3}}{2}\approx0.5-i\,\,\,0.87$

(final values are rounded to 2 decimal places).