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Raising to half power = PRINCIPAL square root?

by perishingtardi
Tags: complex, power, principal, raising, square
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Mar17-14, 12:25 PM
P: 21
This may seem like a very elementary question...but here goes anyway.

When a positive number is raised to the power 1/2, I have always assumed that this is defined as the PRINCIPAL (positive) square root, e.g. [tex]7^{1/2} = \sqrt{7},[/tex]. That is, it does not include both the positive and negative square rootsL [tex]7^{1/2} \neq -\sqrt{7} = -7^{1/2}.[/tex]

In complex analysis, however, this doesn't seem to be the case? E.g. we write [tex](-1)^{1/2} = \pm i.[/tex]

Have I understood these conventions correctly? I have also been thinking about a similar situation: how in real analysis we think of every positive number as having a single natural logarithm, e.g. [tex]\ln 2 = 0.693\dotsc,[/tex] when in fact there are actually infinitely many:
[tex]\ln 2 = 0.693\dotsc + 2\pi n i \qquad (n=0,\pm1,\pm2,\dots).[/tex]
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Mar17-14, 12:29 PM
micromass's Avatar
P: 18,086
In real analysis, you are right to take the principal square root.

In complex analysis, things are a bit more complicated. You have two options here: either you take a multivalues approach where complex exponentiation and logarithms yield not one value but several ones. The other approach is still to take a principal logarithm and a principal exponent. I think the latter approach is more popular in introductory complex analysis texts.

See this too:
Mar17-14, 05:54 PM
Sci Advisor
PF Gold
P: 39,353
The difficulty is that, in the complex numbers, there is no way to unambiguously distinguish a particular root. In the real numbers we can show that, any positive number has, for n even, two nth roots, one positive and the other negative. So we can define the nth principle root to be the positive root. But the complex numbers do not form an ordered field so there is no way to distinguish the roots.

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