Is [math]z^{\frac{1}{2}}[/math] equivalent to the principal square root or the plus-minus square root?

In complex analysis the square root function has two branches. By convention, the principal branch maps the Z plane onto the right hand half of the w plane, allowing for cuts.The other branch is equally valid but has no special name and maps the z plane to the left hand half of the w plane in mirror image of the principal.

It would have to, because z^2/2 is also (z²)^1/2, and (z^a)^b=z^ab. Because 2/2=1, then z^2/2=z¹=z. That is also (z²)^1/2, so therefore it is.

 

Sources:

http://mathworld.wolfram.com/ExponentLaws.html

http://math2.org/math/algebra/exponents.htm

http://tobybartels.name/MATH-0950/2007s/monomials/

http://www.math.hmc.edu/calculus/tutorials/reviewtriglogexp/

The idea of "principal" square root is a matter of convention. Mathematically both square roots are equally valid.

The idea of "principal" square root is a matter of convention. Mathematically both square roots are equally valid.

 

Thanks guys. I was only thinking about a real argument in the square root, but now I also know something about the complex function.

It would have to, because z^2/2 is also (z²)^1/2, and (z^a)^b=z^ab. Because 2/2=1, then z^2/2=z¹=z. That is also (z²)^1/2, so therefore it is.

 

It would have to be what? I can't tell what you're arguing for.