关于利用z^λ主分支计算1^i的推导正确性验证问询

Hey there! Let's walk through your derivation step by step to confirm its accuracy:

• First off, your core definition $z^\lambda = e^{\ln(z) \cdot \lambda}$ is completely correct — this is the standard way to define exponentiation for complex numbers, so you're starting on solid ground.
• Next, your expression for $\ln(z)$ when $z = a + bi$: $\ln(z) = \ln\left(\sqrt{a^2+b2}\right) + i \cdot \arctan\left(\frac{b}{a}\right)$ works perfectly for $z=1$ (since $a=1, b=0$, the arctangent term is straightforward with no domain edge cases to worry about).
• When you get to $\ln(1)$, you noted $\ln(1) = i \cdot k\pi$ for $k=-1,0,1$. A tiny correction here: actually, $k$ can be any integer ($k \in \mathbb{Z}$), not just those three values. That said, when we talk about the principal branch of the logarithm, we restrict the argument $\arg(z)$ to the interval $(-\pi, \pi]$, so for $z=1$, the principal value of $\ln(1)$ is just $0$ (corresponding to $k=0$).
• Your calculation of $1^i = e^{\ln(1) \cdot i}$ leading to $1^i = e^{-k\pi}$ is mathematically sound: multiplying $i \cdot k\pi \cdot i$ gives $-k\pi$, and the exponential simplifies correctly. For the principal branch, this gives $1^i = e^0 = 1$, which is the standard principal value.

Overall, your derivation is mostly correct — the only minor tweak is expanding the possible values of $k$ to all integers, but your focus on the principal branch is spot-on. Great job working through this as a beginner; understanding complex logarithm branches is tricky, and you've nailed the core ideas!

备注：内容来源于stack exchange，提问作者soldaluz