# Complex Analysis: #9 Simple Consequences

**Theorem 17**Assume f, g : G → ℂ are two analytic functions defined on a region G such that the set {z ∈ G : f(z) = g(z)} has an accumulation point. Then f = g.

*Proof*

Let z

_{0}∈ G be such an accumulation point. Then z

_{0}is a zero of the analytic function f − g. But this is not an isolated zero. Therefore f − g = 0, the trivial constant function.

**Theorem 18**Again, f : G → ℂ is analytic and it is not a constant function. Then f(G) is also a region (that is, open and connected) in ℂ.

*Proof*

Since f is continuous, f(G) must be connected. Is f(G) open? Take w

_{0}∈ f(G), and z

_{0}∈ G with f(z

_{0}) = w

_{0}. So then z

_{0}is a zero of the analytic function f − w

_{0}. Since f −w

_{0}is not constant, it follows that z

_{0}is a zero of some particular ﬁnite order. Theorem 16 now shows that w

_{0}lies in the interior of f(G).

**Theorem 19**Let f : G → ℂ be analytic and not constant. (G is a region.) Let z

_{0}∈ G. Then there exists another point z

_{1}with |f(z

_{1})| > |f(z

_{0})|.

*Proof*

For otherwise, f(z

_{0}) would lie on the boundary of f(G) in ℂ, and thus f(G) would not be open in contradiction to theorem 18.

**Theorem 20 (The Lemma of Schwarz)**Let D = {z ∈ ℂ : |z| ≤ 1} be the closed unit disc in ℂ. Assume D ⊂ G, a region in ℂ, and f : G → ℂ is analytic with f(D) ⊂ D and f(0) = 0. Then |f '(0)| ≤ 1 as well, and in fact |f(z)| ≤ |z| for all z ∈ D. If either |f '(0)| = 1 or there exists some z

_{0}with 0 < |z

_{0}| < 1 such that |f(z

_{0})| = |z

_{0}|, then we must have f being a simple rotation. i.e. f(z) = e

^{iθ}· z, for some θ.

*Proof*

Since f(0) = 0, we can write

say, where g is an analytic function, defined in some neighborhood of D. So f '(0) = g'(0) and |f(z)| = |z| · |g(z)| ≤ 1. Thus |g(z)| ≤ 1/|z|. This holds in particular for |z| = 1.

On the other hand, theorem 18 says that g(B(0, 1)) ⊂ ℂ is open in ℂ. [Here B(0, 1) = {z ∈ ℂ : |z| < 1} is the open disc centered at zero, with radius 1.] If there were some point z

_{*}∈ B(0, 1) with |g(z

_{*})| > 1 then we could choose it to be a point such that this value is maximal. However that would then be a boundary point of f(B(0, 1)), contradicting the fact that f(B(0, 1)) is open. Thus |g(z)| ≤ 1 for all z ∈ D. In particular, |f '(0)| = |g(0)| ≤ 1 and |f(z)| = |z| · |g(z)| ≤ |z| for all z ∈ D.

Finally, let us assume that a z

_{0}exists with 0 < |z

_{0}| < 1 such that |f(z

_{0})| = |z

_{0}|. That means |g(z

_{0})| = 1. But according to theorem 19, this can only be true if g is a constant function. Since it is a constant number with absolute value 1, it must be of the form e

^{iθ}, for some θ.

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