Exponents

Work in progress. This note is still being written and incomplete.

Let cCc \in \mathbb{C} and z0z \ne 0.

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

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

Principal Value

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

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

Worked Examples

1+i\sqrt{1+i}

Log(1+i)=12ln2+iπ4\operatorname{Log}(1+i) = \frac{1}{2}\ln 2 + i\frac{\pi}{4} (1+i)1/2=e12Log(1+i)=e14ln2+iπ8=21/4(cosπ8+isinπ8)(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)1/21.0987+0.4551i(1+i)^{1/2} \approx 1.0987 + 0.4551i

iii^i

Logi=iπ2\operatorname{Log} i = i\frac{\pi}{2} ii=eiLogi=eiiπ/2=eπ/20.2079i^i = e^{i\operatorname{Log} i} = e^{i \cdot i\pi/2} = e^{-\pi/2} \approx 0.2079

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

Written by September 13, 2026 1 min read
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