Bloch's theorem (complex variables)

We first prove the case when f(0) = 0, f′(0) = 1, and |f′(z)| ≤ 2 in the unit disk.

By Cauchy's integral formula, we have a bound

${\displaystyle |f''(z)|=\left|{\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f'(w)}{(w-z)^{2}}}\,\mathrm {d} w\right|\leq {\frac {1}{2\pi }}\cdot 2\pi r\sup _{w=\gamma (t)}{\frac {|f'(w)|}{|w-z|^{2}}}\leq {\frac {2}{r}},}$

where γ is the counterclockwise circle of radius r around z, and 0 < r < 1 − |z|.

By Taylor's theorem, for each z in the unit disk, there exists 0 ≤ t ≤ 1 such that f(z) = z + z²f″(tz) / 2.

Thus, if |z| = 1/3 and |w| < 1/6, we have

${\displaystyle |(f(z)-w)-(z-w)|={\frac {1}{2}}|z|^{2}|f''(tz)|\leq {\frac {|z|^{2}}{1-t|z|}}\leq {\frac {|z|^{2}}{1-|z|}}={\frac {1}{6}}<|z|-|w|\leq |z-w|.}$

By Rouché's theorem, the range of f contains the disk of radius 1/6 around 0.

Let D(z₀, r) denote the open disk of radius r around z₀. For an analytic function g : D(z₀, r) → C such that g(z₀) ≠ 0, the case above applied to (g(z₀ + rz) − g(z₀)) / (rg′(0)) implies that the range of g contains D(g(z₀), |g′(0)|r / 6).

For the general case, let f be an analytic function in the unit disk such that |f′(0)| = 1, and z₀ = 0.

• If |f′(z)| ≤ 2|f′(z₀)| for |z − z₀| < 1/4, then by the first case, the range of f contains a disk of radius |f′(z₀)| / 24 = 1/24.
• Otherwise, there exists z₁ such that |z₁ − z₀| < 1/4 and |f′(z₁)| > 2|f′(z₀)|.
• If |f′(z)| ≤ 2|f′(z₁)| for |z − z₁| < 1/8, then by the first case, the range of f contains a disk of radius |f′(z₁)| / 48 > |f′(z₀)| / 24 = 1/24.
• Otherwise, there exists z₂ such that |z₂ − z₁| < 1/8 and |f′(z₂)| > 2|f′(z₁)|.

Repeating this argument, we either find a disk of radius at least 1/24 in the range of f, proving the theorem, or find an infinite sequence (z_n) such that |z_n − z_n−1| < 1/2ⁿ⁺¹ and |f′(z_n)| > 2|f′(z_n−1)|.

In the latter case the sequence is in D(0, 1/2), so f′ is unbounded in D(0, 1/2), a contradiction.

In the proof of Landau's Theorem above, Rouché's theorem implies that not only can we find a disk D of radius at least 1/24 in the range of f, but there is also a small disk D₀ inside the unit disk such that for every w ∈ D there is a unique z ∈ D₀ with f(z) = w. Thus, f is a bijective analytic function from D₀ ∩ f⁻¹(D) to D, so its inverse φ is also analytic by the inverse function theorem.