Exponents

Exponents

Let $c \in \mathbb{C}$ and $z \ne 0$ .

z^c = e^{c\log z}

$z^c$ is multi-valued, since $\log z$ is multi-valued.

Principal Value

Obtained by using $\operatorname{Log} z$ in place of $\log z$ .

z^c = e^{c\operatorname{Log} z}

The principal value $z^c = e^{c\operatorname{Log} z}$ is analytic on $\mathbb{C} \setminus (-\infty, 0]$ and

\frac{\text{d}}{\text{d}z}\,z^c = cz^{c-1}

The domain is larger for some values of $c$ .

$c$ is a non-negative integer
$z^c$ is single-valued and analytic on $\mathbb{C}$ .
$c$ is a negative integer
$z^c$ is single-valued and analytic on $\mathbb{C} \setminus \{0\}$ .
$c$ is any other value
$z^c$ is multi-valued. The negative real axis is the branch cut and $0$ is a branch point.

Solution Reveal Hide

\operatorname{Log}(1+i) = \frac{1}{2}\ln 2 + i\frac{\pi}{4}

(1+i)^{1/2} = e^{\frac{1}{2}\operatorname{Log}(1+i)} = e^{\frac{1}{4}\ln 2 + i\frac{\pi}{8}} = 2^{1/4}\left(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}\right)

Numerically:

(1+i)^{1/2} \approx 1.0987 + 0.4551i

Solution Reveal Hide

\operatorname{Log} i = i\frac{\pi}{2}

i^i = e^{i\operatorname{Log} i} = e^{i \cdot i\pi/2} = e^{-\pi/2} \approx 0.2079

$i^i$ is real for every branch of $\log i$ , since $\log i = i(\pi/2 + 2k\pi)$ for $k \in \mathbb{Z}$ gives $i^i = e^{-(\pi/2+2k\pi)}$ .

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October 17, 2026 1 min read

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