Thursday, January 17, 2013

Complex Analysis: #9 Simple Consequences

  • 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 z0 ∈ G be such an accumulation point. Then z0 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 w0 ∈ f(G), and z0 ∈ G with f(z0) = w0. So then z0 is a zero of the analytic function f − w0. Since f −w0 is not constant, it follows that z0 is a zero of some particular finite order. Theorem 16 now shows that w0 lies in the interior of f(G).


Theorem 19
Let f : G → ℂ be analytic and not constant. (G is a region.) Let z0 ∈ G. Then there exists another point z1 with |f(z1)| > |f(z0)|.

Proof
For otherwise, f(z0) 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 z0 with 0 < |z0| < 1 such that |f(z0)| = |z0|, then we must have f being a simple rotation. i.e. f(z) = e · z, for some θ.

Proof
Since f(0) = 0, we can write

Complex Analysis: #9 Simple Consequences equation pic 1

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 z0 exists with 0 < |z0| < 1 such that |f(z0)| = |z0|. That means |g(z0)| = 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, for some θ.

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