So ef z eref z ec since ef is an entire function

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Unformatted text preview: on a disc of radius π (= distance between 0 and − π , π , which 2 22 are the nearest points where the function is not analytic. One needs to solve cos z = 0 for this ...). 20. Remember that |ez | = eRez (easy to check using the denitions). So ef (z ) = eRef (z ) ≤ ec . Since ef is an entire function we get from Liouville's theorem that ef is constant. Hence ef (z ) = 0 ⇐⇒ f (z ) ef (z ) = 0 ⇐⇒ f (z ) = 0 ⇐⇒ f (z ) constant. We used the fact that ez = 0 for any z . 26.* (It is important to know how to get the relations for the coecients, but nding the exact values of the coecients is challenging in this exercise (exercise 25 is reasonable). You may skip (b) and (c) ) (a) f (z ) − 3f (z ) + 2f (z ) = 0 ...
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This note was uploaded on 12/05/2013 for the course MATH 334 taught by Professor Thind during the Summer '13 term at University of Toronto- Toronto.

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