Question #122815
Simplify (cos2π/3 + isin2π/3)^1/4 and express the result in a form free from trigonometrically expression.
1
Expert's answer
2020-06-25T18:41:23-0400

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

z1n=r1n(cosx+2πkn+isinx+2πkn),k=0,...,n1z^{\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=cos2π3+isin2π3z=cos\,\frac{2\pi}{3}+i\,sin\,\frac{2\pi}{3} we receive:

z14=cos2π3+2πk4+isin2π3+2πk4=cos(π6+πk2)+isin(π6+πk2),k=0,...,3z^{\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:

(z14)0=cosπ6+isinπ6=32+i120.87+i0.5(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

(z14)1=cos2π3+isin2π3=12+i320.5+i0.87(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

(z14)2=cos7π6+isin7π6=32i120.87i0.5(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

(z14)3=cos5π3+isin5π3=12i320.5i0.87(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: (z14)0=32+i120.87+i0.5(z^{\frac{1}{4}})_0=\frac{\sqrt{3}}{2}+i\,\frac{1}{2}\approx0.87\,+i\,\,\,0.5 ; (z14)1=12+i320.5+i0.87(z^{\frac{1}{4}})_1= -\frac{{1}}{2}+i\,\frac{\sqrt{3}}{2}\approx-0.5+i\,\,\,0.87 ; (z14)2=32i120.87i0.5(z^{\frac{1}{4}})_2= -\frac{\sqrt{3}}{2}-i\,\frac{1}{2}\approx-0.87\,-i\,\,\,0.5 ; (z14)3=12i320.5i0.87(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).


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