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

Statement

Let S=C{xR:x0} 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:SC with

exp(F(z))=z  (zS)andF(1)=0,

and it satisfies Log(z)=1/z on S.

Facts & Assumptions

Given: The slit plane S=C{xR:x0}; 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 expL=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 expL=h, then h is nowhere zero and L=h/h (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L4]

For z0 with principal polar form z=r(cosθ+isinθ) and π<θπ, Logz=logr+iθ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers).

[L5]

Every z0 has a unique representation z=r(cosθ+isinθ) with r=z>0 and π<θπ (Every nonzero complex number has a unique polar form r(cosθ+isinθ) with r>0 and π<θπ).

[L6]

For z0 the solutions of expw=z are exactly Logz+2πik for kZ (All logarithms of z0 are Logz+2πik, kZ).

[L7]

For real x,y, exp(x+iy)=ex(cosy+isiny) and exp(x+iy)=ex (exp(x+iy)=ex(cosy+isiny), exp(x+iy)=ex, and eiπ+1=0).

[L9]

A nonempty open URn is star-shaped with respect to aU when a+t(xa)U for every xU and 0t1 (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, logx is the unique real y with expy=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, Rez=a, Imz=b and z=a2+b2 (Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct
1.1

S is open: if z=x+iyS with y0 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 1S.

givenL14L15
1.2

S is star-shaped with respect to 1 in the sense of [L9]: for zS and 0t1 put w=1+t(z1). If w were a real number 0 then Imw=tImz=0 by [L14]; t=0 gives w=1>0, so t>0 and Imz=0, making z a real number, necessarily z>0 because zS; but then w=(1t)+tz>0, a contradiction.

givenL9L14
2.1

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

step 1.1step 1.2L2L10
3.1

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

step 2.1L1L12
4.1

Fix zS and let ϕ(t)=ImF(1+t(z1)) 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(z1) and using exp(F(w))=w together with [L7], [L11] and [L14] gives w=eReF(w)(cos(±π)+isin(±π))=eReF(w), a real number <0, contradicting wS. Hence ImF(z)(π,π).

step 1.2step 3.1L7L8L11L13L14
5.1

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

step 3.1step 4.1L4L5L6L14L16
6.1

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 expF1=h and F1(1)=0, then F1Log is a constant in 2πiZ by [L1], and it vanishes at 1, so F1=Log.

step 5.1L1L3

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