Amaton Posted October 14, 2013 Share Posted October 14, 2013 Is [math]z^{\frac{1}{2}}[/math] equivalent to the principal square root or the plus-minus square root? 1 Link to comment Share on other sites More sharing options...
studiot Posted October 14, 2013 Share Posted October 14, 2013 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. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted October 15, 2013 Share Posted October 15, 2013 It would have to, because z2/2 is also (z2)1/2, and (za)b=zab. Because 2/2=1, then z2/2=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/ Link to comment Share on other sites More sharing options...
mathematic Posted October 15, 2013 Share Posted October 15, 2013 The idea of "principal" square root is a matter of convention. Mathematically both square roots are equally valid. 1 Link to comment Share on other sites More sharing options...
Amaton Posted October 27, 2013 Author Share Posted October 27, 2013 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 z2/2 is also (z2)1/2, and (za)b=zab. Because 2/2=1, then z2/2=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. Link to comment Share on other sites More sharing options...
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