Write 1^(-i) in the form u+iv
Write $1^{-i}$ in the form $u+iv$.
Explanation of Method
As in the previous problem, $1=e^{2k\pi i}$ for $k\in\mathbb Z$, so $1^{-i}=e^{-i\cdot\text{Log}(1)}$ is multi-valued. Here the exponent is purely imaginary, which turns out to make every branch a positive real number.
Step 1 · General logarithm and exponentiation
$$ \text{Log}(1)=2k\pi i, \qquad 1^{-i}=e^{-i(2k\pi i)}=e^{-i^2\cdot2k\pi}=e^{2k\pi} $$
Step 2 · Principal value ($k=0$)
$$ \boxed{1^{-i}\big|_{k=0}=e^0=1} $$
Step 3 · Other branches
$$ k=1:\ \ 1^{-i}=e^{2\pi}\approx535.492, \qquad k=-1:\ \ 1^{-i}=e^{-2\pi}\approx0.001867 $$
Remarkable property
Every value $e^{2k\pi}$ is a positive real number, so $v_k=0$ for all $k$ — the entire (infinite) set of values of $1^{-i}$ lies on the positive real axis, unlike $2^{2i}$ or $1^{\sqrt2}$, whose values spread around the unit circle.
Result
Contrast with an integer exponent like $1^{-6}=1$ (single-valued): an imaginary exponent applied to $1$ instead produces the infinite discrete set $\{\ldots,e^{-2\pi},1,e^{2\pi},\ldots\}$.