# Several values of irrational exponentiation

When talking about a number to a rational exponent, there are as many answers as the denominator of the exponent. Like the question: Is $9^{1/2}$ equal to $3$ or $-3$.

However when we have an irrational exponent like $2^{\sqrt2}$, I cant immediatly see more than one candidate solution: $e^{\sqrt2\log2}$. Are there any ways to interpret irrational exponents such that they give several candidate solutions?

Another way to state this question would be:

are there more than one complex solution to $x^{1/p}=y$ for irrational $p$ and positive real $y$.

• Yes but you really should be thinking of these things as functions of a complex variable. What you want to look up is "branches" of analytic functions. – Gregory Grant May 12 '15 at 19:11
• @GregoryGrant okay, so what would another candidate value of $2^{\sqrt2}$ be? It should obviously satisfy $x^{\sqrt{1/2}}=2$. – Alice Ryhl May 12 '15 at 19:15
• @GregoryGrant What I mean is, if a number is given one irrational number as exponent, can we get several possible values like we did with $\sqrt2$? – Alice Ryhl May 12 '15 at 19:42
• I suppose my question is equivalent to: are there more than one complex solution to $x^{1/p}=y$ for irrational $p$ and positive real $y$. – Alice Ryhl May 12 '15 at 19:51
• Try the complex number $z$ that has argument $2\pi/\sqrt{2}$ and modulus equal to $2$. It lives somewhere in the 3rd quadrant of the complex plane. – Gregory Grant May 12 '15 at 19:59

In the real numbers field we have no ambiguity if we define (as usual) $a^{\frac{1}{2}}$ as the positive root of $a$ and $a^\alpha=e^{\alpha \log a}$ (for $a\le 0$) .
The problems come out when we go to complex numbers, as a consequence of the fact that $e^{i 2 k \pi}=1 \;\forall k \in \mathbb {Z}$. So $y=e^{i x}$ is a periodic function ( also for $x \in \mathbb{R}$) and is not invertible.
So, in the complex numbers field we have: $$y=e^x \iff y=e^x\cdot 1 = e^xe^{i2k\pi}=e^{x+i2k\pi}$$ and the inverse '' function'' is a multivalued function : $$\log y= x+i2k\pi$$ we can find $x$ writing the complex number $y$ in polar form as $y=|y|e^{i\theta}$ so, from $|y|e^{i\theta}=e^x$ we have $x=\log |y| +i\theta$ and: $$\log y= \log |y| +i\theta+i2k\pi$$
As a consequences of this fact for $y,x,\alpha \in \mathbb{R}$ and $y,x >0$ (this imples $\theta=0$), we have: $$y=x^\alpha \iff y=e^{\alpha\log x}=e^{\alpha(\log |x|+i2k\pi)}$$
For $\alpha=1/2$ this gives the two square roots and for $\alpha$ irrational gives, in general, many values.