Find all sixth roots of (-64)

Hi Gibsonj335,

 

 

Find all sixth roots of (-64)

the first one is 2i     this has no real part so it is on the imaginary axis.   it is    \(2e^{i\pi/2}\)

 

the difference in the angles (arguments) is going to be  2pi/6 

 

 

 

the angles will be

 

So the first angle is pi/2 which equals  3pi/6

 

the angles (arguments)  will be

 

 \( \frac{3\pi}{6},\quad \frac{3\pi+2\pi}{6},\quad \frac{3\pi+4\pi}{6},\quad \frac{3\pi+6\pi}{6},\quad \frac{3\pi+8\pi}{6},\quad \frac{3\pi+10\pi}{6}\\~\\ \frac{3\pi}{6},\quad \frac{5\pi}{6},\quad \frac{7\pi}{6},\quad \frac{9\pi}{6},\quad \frac{11\pi}{6},\quad \frac{13\pi}{6}\\~\\ \mbox{so the 6 roots will be:}\\~\\ 2e^{\frac{3\pi i}{6}},\quad 2e^{\frac{5\pi i}{6}},\quad 2e^{\frac{7\pi i}{6}},\quad2e^{ \frac{9\pi i}{6}},\quad 2e^{\frac{11\pi i}{6}},\quad 2e^{\frac{13\pi i}{6}}\\~\\ \)

I wasn't sure of the working so I used this for a reference:

 

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

2 e^((i pi)/6)~~1.7321+1.0000 i  (principal root)

2 i

2 e^((5 i pi)/6)~~-1.7321+1.000 i

-2 i

2 e^(-(5 i pi)/6)~~-1.7321-1.000 i

2 e^(-(i pi)/6)~~1.7321-1.0000 i