# Can raising a number to an irrational power have infinite solutions?

$$a^{\frac{1}{2}}$$ is generally considered to be the positive square root of $$a$$, but it also makes sense (depending on context) to consider it to be multivalued, returning all square roots of $$a$$

Likewise with something like $$a^{\frac{3}{5}}$$- it sometimes makes sense to consider the multivalued function which returns 5 values- the 5 complex values that, when raised to the 5th power, return $$a^3$$.

Extending this idea to irrational numbers, you don't get fractions directly, but you can get them using sequences of rational numbers that converges to the desired irrational. For example, to reach $$\pi$$, you could use 3, 3.1, 3.14, 3.141, 3.1415, etc. For such a sequence, the denominator grows without bound, so raising a number to each subsequent power in the sequence increases the number of complex solutions.

Does it then follow that when considering non-integer exponentiation to be multivalued, raising a number to an irrational power yields an infinite number of complex solutions? If so, would that be every point on a circle concentric with the unit circle, or something else?