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7. Suppose f and g are entire functions and convenience, let p= all z, |f(2)| < 5|g(2)|, this implies that f and g have the same zeros. suppose the has finitely many zeros. For min{|z; – zk| : 9(2;) = g(2k) = 0, z; 2k}. Suppose further that, for a. Show that the only singularities of h =1 are the removable ones at the zeros of g. (Hint: you know something about h in a neighborhood of a zero of g.) b. Prove that there is a constant c, c < 5 so that f(z) = cg(2) for all z. (Hint: use Part a. and an important theorem, applied to an entire function that is usually equal to h.)

7. Suppose f and g are entire functions and convenience, let p= all z, |f(2)| < 5|g(2)|, this implies that f and g have the same zeros. suppose the has finitely many zeros. For min{|z; – zk| : 9(2;) = g(2k) = 0, z; 2k}. Suppose further that, for a. Show that the only singularities of h =1 are the removable ones at the zeros of g. (Hint: you know something about h in a neighborhood of a zero of g.) b. Prove that there is a constant c, c < 5 so that f(z) = cg(2) for all z. (Hint: use Part a. and an important theorem, applied to an entire function that is usually equal to h.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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7. Suppose f and g are entire functions and suppose the g has finitely many zeros. For convenience, let p= all z, |f(z)| < 5[g(z)|, this implies that f and g have the same zeros. min{|2; – zk| : g(2;) = g(zk) = 0, z; 7 zk}. Suppose further that, for a. Show that the only singularities of h I are the removable ones at the zeros of g. (Hint: you know something about h in a neighborhood of a zero of g.) b. Prove that there is a constant c, c| < 5 so that f(2) = cg(2) for all z. (Hint: use Part a. and an important theorem, applied to an entire function that is usually equal to h.)
Transcribed Image Text:7. Suppose f and g are entire functions and suppose the g has finitely many zeros. For convenience, let p= all z, |f(z)| < 5[g(z)|, this implies that f and g have the same zeros. min{|2; – zk| : g(2;) = g(zk) = 0, z; 7 zk}. Suppose further that, for a. Show that the only singularities of h I are the removable ones at the zeros of g. (Hint: you know something about h in a neighborhood of a zero of g.) b. Prove that there is a constant c, c| < 5 so that f(2) = cg(2) for all z. (Hint: use Part a. and an important theorem, applied to an entire function that is usually equal to h.)
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