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The principal logarithm is the normalised holomorphic branch on the slit plane

Statement

Let S=C∖{x∈R:x≤0} be the slit plane. Then S is a complex domain, star-shaped with respect to 1, and homologically simply connected. The principal logarithm Log⁡ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers) is the unique holomorphic F:S→C with

exp⁡(F(z))=z  (z∈S)andF(1)=0,

and it satisfies Log⁡′(z)=1/z on S.

Facts & Assumptions

Given: The slit plane S=C∖{x∈R:x≤0}; segments and star-shapedness in the plane are those of Complex star-shaped and convex domains are the published Euclidean notions under the identification C=R2.

[L1]

On a homologically simply connected complex domain, a holomorphic nowhere-zero f admits a holomorphic L with exp⁡∘L=f, and any two such differ by a constant in 2πiZ (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm).

[L2]

A nonempty open star-shaped subset of C is a complex domain and is homologically simply connected (Star-shaped plane domains are homologically simply connected).

[L3]

If L and h are holomorphic on an open set with exp⁡∘L=h, then h is nowhere zero and L′=h′/h (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L4]

For z≠0 with principal polar form z=r(cos⁡θ+isin⁡θ) and −π<θ≤π, Log⁡z=log⁡r+iθ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers).

[L5]

Every z≠0 has a unique representation z=r(cos⁡θ+isin⁡θ) with r=∣z∣>0 and −π<θ≤π (Every nonzero complex number has a unique polar form r(cos⁡θ+isin⁡θ) with r>0 and −π<θ≤π).

[L6]

For z≠0 the solutions of exp⁡w=z are exactly Log⁡z+2πik for k∈Z (All logarithms of z≠0 are Log⁡z+2πik, k∈Z).

[L7]

For real x,y, exp⁡(x+iy)=ex(cos⁡y+isin⁡y) and ∣exp⁡(x+iy)∣=ex (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0).

[L9]

A nonempty open U⊆Rn is star-shaped with respect to a∈U when a+t(x−a)∈U for every x∈U and 0≤t≤1 (Star-shaped open subsets of Euclidean space).

[L10]

A complex domain is a nonempty, connected, open subset of C (A complex domain is a nonempty connected open subset of C).

[L11]

For x>0, log⁡x is the unique real y with exp⁡y=x (The natural logarithm as the inverse of the exponential function).

[L13]

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

[L14]

For z=a+bi with a,b real, Re⁡z=a, Im⁡z=b and ∣z∣=a2+b2 (Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct
1.1givenL14L15

S is open: if z=x+iy∈S with y≠0 then the ball of radius ∣y∣ about z contains no real number, and if y=0 then x>0 and the ball of radius x about z contains no real number ≤0; in both cases [L14] and [L15] put a ball around z inside S. It is nonempty, since 1∈S.

1.2givenL9L14

S is star-shaped with respect to 1 in the sense of [L9]: for z∈S and 0≤t≤1 put w=1+t(z−1). If w were a real number ≤0 then Im⁡w=tIm⁡z=0 by [L14]; t=0 gives w=1>0, so t>0 and Im⁡z=0, making z a real number, necessarily z>0 because z∈S; but then w=(1−t)+tz>0, a contradiction.

2.1step 1.1step 1.2L2L10

By steps 1.1 and 1.2 and [L2], S is a complex domain and is homologically simply connected.

3.1step 2.1L1L12

The identity function h(z)=z is holomorphic and nowhere zero on S, because 0∉S, so [L1] gives a holomorphic G on S with exp⁡∘G=h; since exp⁡(G(1))=1, [L12] puts G(1) in 2πiZ, and F:=G−G(1) is holomorphic with exp⁡∘F=h and F(1)=0.

4.1step 1.2step 3.1L7L8L11L13L14

Fix z∈S and let ϕ(t)=Im⁡F(1+t(z−1)) for t∈[0,1]; the segment lies in S by step 1.2, and ϕ is continuous by [L13] and [L14], with ϕ(0)=0. If ∣ϕ(1)∣≥π then [L8] gives t with ϕ(t)=π or ϕ(t)=−π; writing w=1+t(z−1) and using exp⁡(F(w))=w together with [L7], [L11] and [L14] gives w=eRe⁡F(w)(cos⁡(±π)+isin⁡(±π))=−eRe⁡F(w), a real number <0, contradicting w∈S. Hence Im⁡F(z)∈(−π,π).

5.1step 3.1step 4.1L4L5L6L14L16

By [L6] there is an integer k with F(z)=Log⁡z+2πik; taking imaginary parts and using [L4] and [L5], Im⁡F(z)=θ+2πk with −π<θ≤π, and step 4.1 gives Im⁡F(z)∈(−π,π), so 2πk∈(−2π,2π) and therefore k=0 by [L16]. Hence F=Log⁡ on S.

6.1step 5.1L1L3∎

By step 5.1 the principal logarithm is holomorphic on S, and [L3] applied to L=Log⁡ and h(z)=z gives Log⁡′(z)=1/z there. If F1 is any holomorphic function on S with exp⁡∘F1=h and F1(1)=0, then F1−Log⁡ is a constant in 2πiZ by [L1], and it vanishes at 1, so F1=Log⁡.

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