What are the five values of (1+i√3)^(3/5)?

AI Thread Summary
The discussion revolves around finding the five values of (1+i√3)^(3/5), specifically the five fifth roots of (1+i√3) raised to the third power. Participants clarify that the task involves solving the equation X^5 = (1+i√3)^3, which equals -8. The significance of -8 is highlighted, as it is a real number, and this leads to a connection with D'Moivre's theorem for calculating the roots. The conversation also touches on the surprising equivalence of (1+i√3) to -2 when raised to the third power. Understanding these roots is essential for solving the given problem accurately.
juliany
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Homework Statement


Find the five values of (1+i√3)^(3/5)

This question was from my recent end of year exam, I hadn't come across a question like it in my revision, does it mean find the five roots of (1+i√3)^(3/5) ?:confused:
 
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You want to find the five 5th roots of (1+i√3)3.
 
you want to find all the solutions of the equation X5 = (1 + i√3)3 = -8.
 
It's remarkable that the third power is equal to -8!

Yes, you can find the five fifth roots of (1+i\sqrt{3})^3 or just apply D'Moivre's theorem to 1+ i\sqrt{3} with n= 3/5.
 
It's remarkable that the third power is equal to -8!

why?
 
Because it's a real number. (-2)3=-8 as well so some might be reluctant enough to say that 1+i\sqrt{3}=-2 :-p
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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