Paul-Eugène Parent Department of Mathematics and Statistics University of Ottawa March 12th, 2014
Outline 1
Holomorphic power Series Proposition Let f (z) = a n (z z o ) n be the holomorphic function defined on B(z o ; R), where R > 0 is the radius of convergence of the corresponding power series. Then f (z) = n=1 na n(z z o ) n 1, and this series has the same radius of convergence. Furthermore, the coefficients a n are given by a n = f (n) (z o ). n!
Proof By the ACT, f (z) = n=1 na n(z z o ) n 1 on B(z o ; R). Hence its radius of convergence is at least R. Suppose there is z 1 C such that z 1 z o > R and n=1 na n(z 1 z o ) n 1 converges. Then necessarily the sequence na n ro n is bounded, where r o = z 1 z o (why?). Thus a n ro n = (na n ro n 1 ) r o n is also bounded. This, by Abel-Weierstrass, implies that a n(z z o ) n converges for all z D(z o ; r) and all R r < r o. This contradicts the maximality of R.
Conclusion To identify the coefficients we proceed inductively. Clearly f (z o ) = a n(z o z o ) n = a 0. Moreover, f (n) (z) = n!a n + k=n+1 k(k 1)(k 2)...(k n+1)(z z o ) k n, and setting z = z o, we get f (n) (z o ) = n!a n. Corollary If a n(z z o ) n = b n(z z o ) n on some B(z o ; ρ), then a n = b n for all n N.
Exercises Proposition Consider the power series a n(z z o ) n. Ratio test: If lim n a n a n+1 exists, then it equales R, the radius of convergence of the series. Root test (Hadamard s formula): Consider ρ = lim sup n a n (which always exists). Then R = 1 ρ.
Theorem Let f : A C be holomorphic on an open set A C. Let z o A and choose B(z o ; ρ) A. Then for every z B(z o ; ρ), the series f (n) (z o ) (z z o ) n n! converges (hence the radius of convergence of the series is at least ρ). Moreover, for all z B(z o ; ρ), f (z) = f (n) (z o ) (z z o ) n. n!
Remarks: The series is called the Taylor series of f around the point z o. In the real case, a C function f : A R is called analytic if for all x o A f (x) = f (n) (x o ) (x x o ) n n! on some non-trivial (x o ρ, x o + ρ) A. In the complex case, by Taylor s theorem, to be analytic on A is equivalent to be holomorphic on A.
Exercise Consider the real-valued function f : R R { e 1/x 2 x 0. x 0 x = 0 Show that it is C everywhere but fails to be analytic at x = 0. If you replace x by z C in the formula above, what can you say about the associated complex-valued function at z = 0?
Proof of Exercise: Show that zn converges uniformly on all D(0; r) B(0; 1) towards 1/(1 z). Let 0 < σ < ρ, γ(t) = σe 2πit + z o, and pick z B(z o ; σ). Then Cauchy s Integral formula gives us f (z) = 1 f (ζ) 2πi ζ z. By construction we have z z o ζ z o < 1. Hence... γ
1 ζ z = = 1 1 ζ z o 1 z zo ζ z o 1 ( ) z n zo, ζ z o ζ z o which converges uniformly on γ([0, 1]) (exercise). We can now write f (z) = 1 [ f (ζ) ( ) ] z n zo 2πi γ ζ z o ζ z o = 1 [ ] f (ζ)(z z o ) n 2πi (ζ z o ) n+1, γ
... where we notice that multiplying a uniformly convergence series by a bounded function, the convergence remains uniform. Hence we can interchange the integral and the series to obtain f (z) = = = 1 2πi [ γ f (ζ)(z z o ) n (ζ z o ) n+1 [ (z z o ) n 1 2πi γ (z z o ) n f (n) (z o ) n! ] f (ζ) (ζ z o ) n+1 ], i.e., Cauchy s Integral formula for higher derivatives gives us the desired result.
A first consequence Let f : A C be an analytic function on an open set A C. Two things can happen at z o A: either f (n) (z o ) = 0 for all n N which implies that f 0 on some neighborhood of z o ; or there is a smallest natural number n such that f (n) (z o ) 0. In this case, in a neighborhood of z o f (z) = a k (z z o ) k k=n = (z z o ) n a k+n (z z o ) k. k=0 } {{ } =:φ(z)
In the second case we say that f admits a zero of order n at z o A. Moreover, we see that the series φ(z) has the same radius of convergence as the original series and that φ(z o ) = a n 0. By continuity, φ is none zero in some neighborhood of z o, which implies Corollary If an analytic function f admits a zero at z o, then either that zero is isolated or the function is identically zero in a neighborhood of z o.