Since 0 6∈ Ω, 1/z is holomorphic on Ω

1. Complex Logarithm

Theorem 1.1. Let Ω be a simply connected domain in C with 1 ∈ Ω and 0 6∈ Ω. Then there exists a unique holomorphic function F (z) on Ω such that

(1) F (1) = 0 and F⁰(z) = 1/z on Ω.

(2) e^{F (z)}= z for all z ∈ Ω

(3) F (r) = ln r when r is a positive real number close to 1.

Proof. Since 0 6∈ Ω, 1/z is holomorphic on Ω. Let γ be a piecewise smooth curve connecting 1 and z. Define

F (z) = Z

γ

dw w .

Since Ω is simply connected and 1/z is holomorphic on Ω, F (z) is independent of choice of γ. Hence we write

F (z) = Z z

1

dw

w , z ∈ Ω.

Then F defines a function on Ω with F (1) = 0. It is easy to verify that F (z) is holomorphic such that F⁰(z) = 1/z on Ω. To prove (2), we let

g(z) = ze^{−F (z)}, z ∈ Ω.

Then g(z) is holomorphic on Ω with g(1) = 1. Then

g⁰(z) = e^{−F (z)}+ z · e^{−F (z)}· (−F⁰(z))

= e^{−F (z)}− z · e^{−F (z)}·1 z

= e^{−F (z)}− e^{−F (z)}= 0.

Since Ω is a simply connected domain and g⁰(z) = 0 on Ω, g(z) is a constant function. Thus g(z) = g(1) = 1 for all z ∈ Ω.

When r is real and close to 1, then

F (r) = Z r

1

dw w .

Let us choose the path from 1 to r on the axis defined by w(t) = t for 1 ≤ t ≤ r if r ≥ 1, or r ≤ t ≤ 1 if r < 1. Then

F (t) = Z _r

1

dt

t = ln r.

Notice that such a function is unique by the coincidence principle.  We see that F (z) is an extension of the standard logarithm. We call F (z) a branch of log z and denote F (z) by log_Ωz. For example, Ω = C \ (−∞, 0] is a simply connected domain in C such that 0 6∈ Ω and 1 ∈ Ω. Then there is a unique holomorphic function F on Ω satisfying (1)-(3). This function is defined by the complex integration:

F (z) = Z z

1

dw w .

Suppose z = re^θ for −π < θ < π. Without loss of generality, we assume r > 1. Let L₁ be the path from 1 to r defined by w(t) = t for 1 ≤ t ≤ r and L2 be the arc w(t) = re^it for

1

2

0 ≤ t ≤ θ. Then the path integral is equal to F (z) =

Z

L1

dw w +

Z

L2

dw w

= Z r

1

dt t +

Z θ 0

rie^it e^it dt

= ln r + iθ.

Corollary 1.1. Let Ω = C \ (−∞, 0] and F (z) be the branched of log z in Ω. If z = re^iθ for r > 0 and −π < θ < π, then

F (z) = ln r + iθ.

Definition 1.1. The branch of log z defined on Ω is called the principal branch of log z.

In calculus, we have already learned that ln(1 + x) has the following series expansion:

ln(1 + x) = x − x² 2 +x³

3 + · · · =

∞

X

n=1

(−1)ⁿxⁿ

n , −1 < x < 1.

Let F (z) be the principal branch of log z. We will prove that F (1 + z) =

∞

X

n=1

(−1)ⁿzⁿ

n, |z| < 1.

Let G(z) be the power series P∞

n=1(−1)ⁿzⁿ/n defined on D = {z : |z| < 1}. Then G(z) is holomorphic on D For any 0 < r < 1, we have

F (1 + r) = ln(1 + r) = G(r).

We find F (1 + z) = G(z) on [0, 1]. Since D \ (−∞, 0] is connected and F (1 + z) = G(z) on [0, 1], by coincidence principle, F (1 + z) = G(z) for all D \ (−∞, 0].

Theorem 1.2. Let f be a nowhere vanishing holomorphic function on a simply connected domain Ω. Then there exists a holomorphic function g on Ω such that

f (z) = e^g(z).

In this case, we denote g(z) by log f (z). It determine a branch of log z.

Proof. Given z₀∈ Ω, let c₀ be a constant such that e^c0 = f (z₀). Define g(z) = c₀+

Z z z0

f⁰(w) f (w)dw.

Since f is nonzero on Ω, f⁰/f is holomorphic on Ω. Since Ω is simply connected, the path integral is independent of choice of paths from z0 to z. Moreover, g is holomorphic on Ω such that g⁰(z) = f⁰(z)/f (z). Define a holomorphic function G(z) on Ω by

G(z) = f (z)e^−g(z).

Then G⁰(z) = 0 on Ω. Since Ω is simply connected, G is a constant function. Since G(z0) = f (z₀)e^−c0 = 1, we see that G(z) = 1 on Ω. Thus f (z) = e^g(z) on Ω. 

Let y0 ∈ R and denote

A_y0 = {x + iy : x ∈ R, y0≤ y < y₀+ 2π}.

Then one can show that the function

exp : Ay0 → C \ {0}

3

is a bijection. In other word, we may define log_Ay0 to be the inverse of exp : Ay0 → C \ {0}.

Thus log_Ay0 is a branch of log z. When y0 = −π, we obtain the (extension of) principal branch of log z.

Proposition 1.1. Choose a branch of log z. For any z1, z2 ∈ C \ {0}, log(z₁· z₂) = log z₁+ log z₂ mod 2π.

After choosing a branch of log z, we define z^α= e^{α log z} for any α ∈ C. The n-th root function √ⁿ

z is defined by

√n

z = zⁿ¹

when a branch of log z is chosen. In this case, we call √ⁿ

z is a branch of the n-th root function.

Proposition 1.2. Let −π < θ < π and z = re^iθ. Choose the principal branch for log z.

Then

√n

z =√ re^iθn.

Proof. Let log z be the principal branch of log . Then log z = r + iθ. Thus

√n

z = e^{n1log z} = e^{ln r+iθn} = e^{ln rn} e^θni = √ⁿ re^iθn.