1. Mar 23, 2005

### saltydog

When I use Solve to solve for the roots of $x^6+1=0$, Mathematica returns the following roots:

$$i$$
$$-i$$
$$(-1)^{\frac{1}{6}$$
$$-(-1)^{\frac{1}{6}}$$
$$(-1)^{\frac{5}{6}}$$
$$-(-1)^{\frac{5}{6}}$$

I realize these are derived from the nth roots of unity but I don't understand how Mathematica is assigning the various roots to the various fractional powers of 1/6 and 5/6. For example, if I had used $x^7$, then Mathematica returns values of -1 raised to 4/7, 5/7, and 6/7. How do I know what n-th root is being assigned to the power of 4/7 for example. I know I can evaluate it via N[] but I'd like to know the assignment scheme.

Anyone know?

2. Jul 7, 2005

### lurflurf

they are assigned by principle value. Like sqrt(4) means 2 normally. It is easier to when looking at the numbers in polar form where the angle is restricted to (-pi,pi]
thus
z=|z|exp(iArctan(Im(z)/Re(z))=r*exp(i a)=(r<a)
$$i$$
=(1<pi/2)
$$-i$$
=(1<-pi/2)
$$(-1)^{\frac{1}{6}$$
=(1<pi)^(1/6)
=(1<pi/6)
$$-(-1)^{\frac{1}{6}}$$
=-(1<pi/6)
=(1<-5pi/5)
$$(-1)^{\frac{5}{6}}$$
=(1<pi)^(5/6)
=(1<5pi/6)
$$-(-1)^{\frac{5}{6}}$$
=-(1<5pi/6)
=(1<-pi/6)
in other words to find the priciple value of z^x for z complex and x real
write z=(r<a) with -pi<a<=pi
z^x=(r^a<x*a)
where x*a is reduced back into (-pi,pi] if needed.

note this is not something mathematica does just because. It is the standard convention for principle values of roots. That's why have the high school teachers who say the principle value cube root of -1 is -1 should be dunked in rotten cabage.

Last edited: Jul 7, 2005