Write 1^(sqrt2) in the form u+iv
Write $1^{\sqrt2}$ in the form $u+iv$.
Explanation of Method
Since $1=e^{i\cdot2k\pi}$ for any integer $k$, the general complex power $z^\alpha=e^{\alpha\,\text{Log}(z)}$ is multi-valued whenever $\alpha$ is not an integer. Compute the general family, then isolate the principal value ($k=0$).
Step 1 · General logarithm of $1$
$$ \text{Log}(1)=\ln1+i(0+2k\pi)=2k\pi i, \qquad k\in\mathbb Z $$
Step 2 · Raise to the power $\sqrt2$
$$ 1^{\sqrt2}=e^{\sqrt2\cdot2k\pi i}=\cos(2k\pi\sqrt2)+i\sin(2k\pi\sqrt2) $$
Step 3 · Principal value ($k=0$)
$$ \boxed{1^{\sqrt2}\big|_{k=0}=\cos0+i\sin0=1} $$
Step 4 · Other branches ($k\neq0$)
For $k=1$: $\ 2\pi\sqrt2\approx8.885766$ rad, so:
$$ 1^{\sqrt2}\big|_{k=1}=\cos(8.885766)+i\sin(8.885766)\approx-0.8582+0.5133i $$
For $k=-1$ (using $\cos(-x)=\cos x,\ \sin(-x)=-\sin x$):
$$ 1^{\sqrt2}\big|_{k=-1}\approx-0.8582-0.5133i $$
Verification
Since $\sqrt2$ is irrational, $2k\pi\sqrt2$ is never a multiple of $2\pi$ for $k\neq0$, so all branches give genuinely distinct values — the set $\{1^{\sqrt2}\}$ is infinite and dense on the unit circle.
Result
Irrationality of the exponent's interaction with $2\pi$ is exactly what keeps every branch distinct here — with a rational exponent, only finitely many values would occur.