Multivariable Advanced Calculus

14.7. EXERCISES 411Now explain why you can switch the sum and the integral. You will need to arguethe sum converges uniformly which is what will justify this manipulation. Nextuse the result of Theorem 14.6.11.14. Prove the following amazing result about the zeros of an analytic function. LetΩ be a connected open set (region) and let f : Ω → X be analytic. Then thefollowing are equivalent.(a) f (z) = 0 for all z ∈ Ω(b) There exists z 0 ∈ Ω such that f (n) (z 0 ) = 0 for all n.(c) There exists z 0 ∈ Ω which is a limit point of the set,Z ≡ {z ∈ Ω : f (z) = 0} .Hint: From Problem 13, if c.) holds, then for z near z 0∞∑ f (n) (z 0 )f (z) =(z − z 0 ) nn!Say f (n) (z 0 ) ≠ 0. Then considern=mf (z)(z − z 0 ) m = f (m) (z 0 )m!+∞∑n=m+1f (n) (z 0 )n!(z − z 0 ) n−mNow let z n → z 0 , z n ≠ z 0 but f (z n ) = 0. What does this say about f (m) (z 0 )?Clearly the first two conditions are equivalent and they imply the third.15. You want to define e z for z complex such that it is analytic on C. Using Problem14 explain why there is at most one way to do it and still have it coincide with e xwhen z = x + i0. Then show using the Cauchy Riemann equations thate z ≡ e x (cos (y) + i sin (y))is analytic and agrees with e x when z = x + i0. Also showddz ez = e z .Hint: For the first part, suppose two functions, f, g work. Then consider f − g.this is analytic and has a zero set, R.16. Do the same thing as Problem 15 for sin (z) , cos (z) . Also explain with a very shortargument why all identities for these functions continue to hold for the extendedfunctions. This argument shouldn’t require any computations at all. Why issin (z) no longer bounded if z is allowed to be complex? Hint: You might trysomething involving the above formula for e z to get the definition.17. Show that if f is analytic on C and f ′ (z) = 0 for all z, then f (z) ≡ c for someconstant c ∈ C. You might want to use Problem 14 to do this really quickly. Nowusing Theorem 14.6.11 prove Liouville’s theorem which states that a functionwhich is analytic on all of C which is also bounded is constant. Hint: By thattheorem,f ′ (z) = 1 ∫2πi C rf (w)(w − z) 2 dwwhere C r is the positively oriented circle of radius r which is centered at z. Nowconsider what happens as r → ∞. You might use the corresponding version ofTheorem 14.2.4 applied to contour integrals and note the total length of C r is 2πr.