Is $1$ raised to any complex power equal to $1$?
I just saw a solution that says it is since, for any complex number $z$
$$1^z = e^{z\log1} = e^{z(0)} = 1$$
However, isn't this only true for the principle branch of $1^z$, since by definiton, (letting capital L denote the princple value of log
$$\log1 = \operatorname{Log}1 + i2\pi k$$
for any $k \in \mathbb{Z}$?
$\endgroup$ 32 Answers
$\begingroup$You are correct: $1$ is not the only value of $1^z$ if $z$ is not an integer. For example, $-1$ is also one of the values of $1^{1/2}$. In general the possible values of $1^z$ are $e^{2 \pi i n z}$ for integers $n$.
$\endgroup$ $\begingroup$$1.~~$ Euler's identity: $$e^{i~\pi}+1=0 \implies e^{i~\pi} = -1 \implies (e^{i~\pi})^i = (-1)^i = e^ {-\pi}\tag1$$
$2.~~$ $$e^{i~\pi}+1=0 \implies 1^i = (- e^{i~\pi})^i \implies 1^i = (-1)^i~(e^{i~\pi})^i) = = e^{-\pi}~e^{-\pi} \implies 1^i = e^{-2~\pi}$$
Hence $~1~$ raised to the power of $~i~$ equals $~-0.0018674427\cdots~$
Strange fact or wrong?
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