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Different branches shift logarithms by $2 π i k$ and complex powers by exponential factors

Statement

Let $V \subseteq C$ be a connected open set with $0 \notin V$ , and let $L_{1}, L_{2} : V \to C$ be two holomorphic logarithm branches of $z$ on $V$ . There is a unique integer $k \in Z$ with

$L_{1} (z) - L_{2} (z) = 2 π i k for every z \in V .$

Consequently, for every $α \in C$ the branch powers of Complex powers defined from a holomorphic logarithm branch differ by

$z_{L_{1}}^{α} = e^{2 π i α k} z_{L_{2}}^{α} (z \in V) .$

Additive and multiplicative branch laws are subject to exactly this discrepancy: a claimed identity $\log (z w) = \log z + \log w$ or $(z^{α})^{β} = z^{α β}$ between branch values holds only after the relevant discrepancy vanishes on the points involved, and the companion page exhibits the principal-branch failures.

Facts & Assumptions

Given: A connected open $V \subseteq C$ with $0 \notin V$ , and holomorphic logarithm branches $L_{1}, L_{2}$ of $z$ on $V$ ; $α \in C$ .

[F1]

A holomorphic logarithm branch $L$ of $z$ on $V$ satisfies $\exp (L (z)) = z$ for every $z \in V$ , and its branch power is $z_{L}^{α} : = \exp (α L (z))$ (Complex powers defined from a holomorphic logarithm branch).

[F2]

$\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[F3]

For all $u, v \in C$ , $\exp (u + v) = \exp u \exp v$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[F4]

A continuous image of a connected space is connected (A continuous image of a connected space is connected, and connectedness is a topological property).

[F5]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1F1F2given

[F1] gives $\exp (L_{1} (z)) = z = \exp (L_{2} (z))$ ; by [F2], $L_{1} (z) - L_{2} (z) \in 2 π i Z$ .

2.1F4F5step 1.1

$L_{1} - L_{2}$ is continuous by [F5], so [F4] makes its image connected; step 1.1 gives one integer $k$ with $L_{1} - L_{2} \equiv 2 π i k$ on $V$ .

3.1F1F3step 2.1algebra∎

Substituting step 2.1 into [F1]: $z_{L_{1}}^{α} = \exp (α L_{2} (z) + 2 π i α k) = e^{2 π i α k} \exp (α L_{2} (z)) = e^{2 π i α k} z_{L_{2}}^{α}$ , using [F3] in the middle equality.

Depends on

Used by

Continuing a square root once around the origin changes its sign Example

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

Different branches shift logarithms by 2πik2πik and complex powers by exponential factors