Calculate |(-i)^(-i)| (principal value)

Calculate $\left|(-i)^{-i}\right|$ (principal value).

Explanation of Method

Use $|z^w|=e^{\text{Re}(w\,\text{Log}\,z)}$, which follows directly from $z^w=e^{w\,\text{Log}\,z}$ and $|e^\zeta|=e^{\text{Re}\,\zeta}$.

Step 1 · Polar form of $-i$

$$ -i=e^{-i\pi/2}, \qquad \text{Log}(-i)=\ln1-i\frac\pi2=-i\frac\pi2 $$

Step 2 · Multiply by the exponent $-i$

$$ (-i)\cdot\left(-i\frac\pi2\right)=i^2\cdot\frac\pi2=-\frac\pi2 $$

This product is purely real.

Step 3 · Exponentiate

$$ (-i)^{-i}=e^{-\pi/2} $$

Since the exponent $-\pi/2$ turned out real, $(-i)^{-i}$ itself is a positive real number.

Step 4 · Take the modulus

$$ \boxed{\left|(-i)^{-i}\right|=e^{-\pi/2}\approx0.207880} $$

Verification

By the general formula, $|z^w|=e^{\text{Re}(w\,\text{Log}\,z)}=e^{\text{Re}(-\pi/2)}=e^{-\pi/2}$, matching Step 3 exactly. Coincidentally this equals the principal value of $i^i$ found earlier — a genuine numerical coincidence, not a general identity, since $i$ and $-i$ are different bases.

Result

$$ \left|(-i)^{-i}\right|=e^{-\pi/2}=\frac1{e^{\pi/2}}\approx0.2079 $$

Whenever $w\,\text{Log}(z)$ is real, $z^w$ itself is real and its modulus is just $|z^w|$ itself — no separate argument computation is needed.