Is $2^\pi$ always equal to 8.82498...?

Question

A complex power is defined as:

$z^w = e^{w \ln(z)}$

Since $\ln(z)$ is multi-valued:

$z^w = e^{w(\ln|z| + i(\theta + 2k\pi))}, \quad k \in \mathbb{Z}$

In this case, $\theta = 0$ because $z = 2$ is a positive real number, and also $w = \pi$, so:

$2^\pi = e^{\pi(\ln|2| + i \cdot 2k\pi)}, \quad k \in \mathbb{Z}$

$2^\pi = e^{\pi \ln|2|} \cdot e^{i \cdot 2k \cdot \pi^2}$

Using the properties of logarithms and Euler's identity:

$2^\pi = |2|^\pi \cdot (\cos(2k\pi^2) + i \sin(2k\pi^2))$

Which will lead to infinite different values, but the principal value is with $k = 0$:

$2^\pi \approx |2|^\pi \approx 8.82498$

But another possible value is for example with $k = 1$:

$2^\pi \approx |2|^\pi \cdot (\cos(2\pi^2) + i \sin(2\pi^2)) \approx 5.55693 + i \cdot 6.85571$

Am I right or did I misunderstand something?

Answer 1

The number $2^{\pi}$ is by common convention the real number only. We can read in Complex Analysis by L. Ahlfors in section 3.4 The Logarithm:

(L. Ahlfors): By convention the logarithm of a positive number shall always mean the real logarithm, unless the contrary is stated. The symbol $a^b$, where $a$ and $b$ are arbitrary complex numbers except for the condition $a\ne 0$, is always interpreted as an equivalent of $\exp(b\log a)$. If $a$ is restricted to positive numbers, $\log a$ shall be real, and $a^b$ has a single value. Otherwise $\log a$ is the complex logarithm, and $a^b$ has in general infinitely many values which differ by factors $e^{2\pi i n b}$.