Answer to Question #120442 in Complex Analysis for anav

Question #120442
z^3=6 ( cos ( π/3 ) + i sin ( π/63 ) )
1
Expert's answer
2020-06-08T19:05:16-0400

According to the De Moivre's Formula, all n-th roots of a complex number "r(\\cos(\\theta)+i\\sin(\\theta))"are given by "\\sqrt[n]{r}\\big(\\cos({\\theta+2\\pi k\\over n})+i\\sin({\\theta+2\\pi k\\over n})\\big)," "k=0,1,2,...,n-1"

We have that "r=6, \\theta=\\dfrac{\\pi}{3}, n=3."

"k=0:"


"z_1=\\sqrt[3]{6}\\big(\\cos({{\\pi\\over 3}+0\\over 3})+i\\sin({{\\pi\\over 3}+0\\over 3})\\big)=""=\\sqrt[3]{6}\\big(\\cos({\\pi\\over 9})+i\\sin({\\pi\\over 9})\\big)=""=\\sqrt[3]{6}\\cos({\\pi\\over 9})+i\\sqrt[3]{6}\\sin({\\pi\\over 9})"

"k=1:"


"z_2=\\sqrt[3]{6}\\big(\\cos({{\\pi\\over 3}+2\\pi\\over 3})+i\\sin({{\\pi\\over 3}+2\\pi\\over 3})\\big)=""=\\sqrt[3]{6}\\big(\\cos({7\\pi\\over 9})+i\\sin({7\\pi\\over 9})\\big)=""=-\\sqrt[3]{6}\\cos({2\\pi\\over 9})+i\\sqrt[3]{6}\\sin({2\\pi\\over 9})"

"k=2:"


"z_3=\\sqrt[3]{6}\\big(\\cos({{\\pi\\over 3}+4\\pi\\over 3})+i\\sin({{\\pi\\over 3}+4\\pi\\over 3})\\big)=""=\\sqrt[3]{6}\\big(\\cos({13\\pi\\over 9})+i\\sin({13\\pi\\over 9})\\big)=""=-\\sqrt[3]{6}\\cos({4\\pi\\over 9})-i\\sqrt[3]{6}\\sin({4\\pi\\over 9})"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS