I thought there was this rule that $e^{xB}=(e^x)^B$?

Also what I don't understand is that $x^{\frac{1}{2}}$ is defined as the square root of $x$. And because the square root may also be in $\mathbb{C}$ it doesn't matter which number $x$ is because every complex number has a n-th root.

But for $1^{\frac{1}{2}}$ we have the roots $-1$ and $1$, so I could also write $-1$ on the right side then the Statement above would be true and not true at the same time.

But we said the Expression in the Question is correct and to write $-1$ is false. Can somebody explain the reason we choose one Version over another altough they are equivalent?

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    $\begingroup$ To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots. $\endgroup$ – mrtaurho Jan 27 at 12:07
  • $\begingroup$ @mrtraurho Well, $0$ doesn't have distinct $n$th roots. $\endgroup$ – J.G. Jan 27 at 12:08
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    $\begingroup$ Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised? $\endgroup$ – Did Jan 27 at 12:10
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    $\begingroup$ The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=\sqrt{25}=-5$ which is a contradiction as well. $\endgroup$ – mrtaurho Jan 27 at 12:13
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    $\begingroup$ Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add). $\endgroup$ – Did Jan 27 at 12:18

If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $\sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has $$ 1^{1/2}=1. $$

If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.

The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.

To quote Wikipedia:

Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.

See Failure of power and logarithm identities for more examples.


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