Class 1/28 1 Zeros of an Analytic Function

Class 1/28 1 Zeros of an Analytic Function

<p></p><ul style="display: flex;"><li style="flex:1">Math 752 </li><li style="flex:1">Spring 2011 </li></ul><p></p><p>Class 1/28 <br>1 Zeros&nbsp;of an analytic function </p><p>Towards the fundamental theorem of algebra and its statement for analytic functions. </p><p>Definition 1. Let f : G → C be analytic and f(a) = 0.&nbsp;a is said to have multiplicity m ≥ 1 if there exists an analytic function g : G → C with g(a) = 0 so that </p><p>f(z) = (z − a)<sup style="top: -0.3753em;">m</sup>g(z). </p><p>Definition 2. If f is analytic in C it is called entire. <br>An entire function has a power series expansion with infinite radius of convergence. </p><p>Theorem 1 (Liouville’s Theorem). If f is a bounded entire function then f is constant. </p><p>Proof. Assume |f(z)| ≤ M for all z ∈ C. Use&nbsp;Cauchy’s estimate for f<sup style="top: -0.3299em;">0 </sup>to obtain that |f<sup style="top: -0.3299em;">0</sup>(z)| ≤ M/R for every R &gt; 0 and hence equal to 0. </p><p>Theorem 2 (Fundamental theorem of algebra). For every non-constant </p><p>polynomial there exists a ∈ C with p(a) = 0. </p><p>Proof. Two facts: If p has degree ≥ 1 then lim p(z) = ∞ </p><p>z→∞ </p><p>where the limit is taken along any path to ∞ in C<sub style="top: 0.1364em;">∞</sub>. (Sometimes&nbsp;also written as |z| → ∞.) If p has no zero, its reciprocal is therefore entire and bounded. Invoke Liouville’s theorem. </p><p>Corollary 1. If p is a polynomial with zeros a<sub style="top: 0.1363em;">j </sub>(multiplicity k<sub style="top: 0.1363em;">j</sub>) then p(z) = </p><p>c(z − a<sub style="top: 0.1364em;">1</sub>)<sup style="top: -0.3299em;">k</sup><sup style="top: 0.0922em;">1 </sup>(z − a<sub style="top: 0.1363em;">2</sub>)<sup style="top: -0.3299em;">k</sup><sup style="top: 0.0922em;">2 </sup>...(z − a<sub style="top: 0.1363em;">m</sub>)<sup style="top: -0.3299em;">k</sup><sup style="top: 0.083em;">m </sup></p><p>.</p><p>Proof. Induction, and the fact that p(z)/(z − a) is a polynomial of degree n − 1 if p(a) = 0. </p><p>1<br>The zero function is the only analytic function that has a zero of infinite </p><p>2</p><p>order. (Compare with the real situation, e.g, x → e<sup style="top: -0.3299em;">−1/x </sup>at the origin.) </p><p>Theorem 3. G connected open set, f : G → C analytic. Equivalent: <br>1. f ≡ 0, 2. there is a ∈ G with f<sup style="top: -0.3298em;">(n)</sup>(a) = 0 for all n ≥ 0, 3. the zeros of f in G have a limit point. </p><p>Proof. The first statement implies the other two.&nbsp;We show that the third implies the second: Assume that the zeros of f have a limit point a and that f<sup style="top: -0.3299em;">(n)</sup>(a) = 0 for some n. Let&nbsp;R &gt; 0 be such that f is analytic in a ball of radius R about a. <br>Then define g by </p><p>f(z) = (z − a)<sup style="top: -0.3753em;">n−1</sup>g(z), </p><p>(1) and use power series expansions to show that g is analytic in this ball with g(a) = 0.&nbsp;Since a is a limit point of zeros, (1) shows that g(a) = 0, a contradiction. <br>Note that the second statement implies the first in a disk right away. <br>(Power series expansions!).&nbsp;For general open sets, consider A = {z ∈ G : f<sup style="top: -0.3298em;">(n)</sup>(z) = 0 for all n}. Every&nbsp;limit point of A in G lies in A (since the derivatives are continuous). But A is also open in G: For z ∈ A there exists a disk with center z in G, and a power series expansion shows that f is the zero function in the disk.&nbsp;Hence the disk is a subset of A. It&nbsp;follows that A = G. </p><p>Corollary 2. f ≡ g in an open, connected G iff the set of points with f(z) = g(z) has a limit point. </p><p>Corollary 3. f analytic and not identically zero in an open, connected G, and a ∈ G with f(a) = 0. Then&nbsp;there exists integer n ≥ 1 and analytic </p><p>g : G → C with g(a) = 0 and f(z) = (z − a)<sup style="top: -0.3299em;">n</sup>g(z). Proof. Let n be so that f<sup style="top: -0.3299em;">(k)</sup>(a) = 0 for k &lt; n and f<sup style="top: -0.3299em;">(n)</sup>(a) = 0.&nbsp;Define g(z) = (z − a)<sup style="top: -0.3299em;">−n</sup>f(z) and note that g<sup style="top: -0.3299em;">0</sup>(z) exists for all z ∈ G\{a}. Use&nbsp;a power series expansion of f to show that g is analytic at a as well. </p><p>Corollary 4. For every zero a of an analytic f there exists a punctured disk with center a on which f is non-zero. </p><p>2</p><p>Theorem 4 (Maximum Modulus Theorem). If G is a region (open, connected subset of C) and f : G → C analytic such that there exists a ∈ G with |f(a)| ≥ |f(z)| for all z ∈ G, then f is constant. </p><p>The proof will be given next time. <br>3</p>

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