Math 425 Fall All About Zero

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1 Math 425 Fall 2005 All About Zero These notes supplement the discussion of zeros of analytic functions presented in 2.4 of our text, pp Throughout: Unless stated otherwise, f is a function analytic on a domain D of the plane, z 0 is a point of D, and is the largest open disc in D centered at z 0. We ve seen previously that: 1. Zeros and Power Series f can be represented by a power series with center at z 0 : (1) f(z) = a n (z z 0 ) n valid for all z, n=0 We ve also noted that by repeatedly differentiating both sides of (1) above (using the term-by-term differentiation theorem on the right), and evaluating at z 0 we obtain the coefficients a n in terms of the derivatives of f at z 0 : (2) a n = f (n) (z 0 ) n! for n = 0, 1, 2,..., where the conventions for n = 0 are: f (0) (z 0 ) = f(z 0 ), and 0! = 1. 1 In particular, f(z 0 ) = a 0, hence: (3) f(z 0 ) = 0 a 0 = 0 f(z) = a n (z z 0 ) n, valid in Theorem. If f(z 0 ) = 0 then f(z) = (z z 0 )g(z) for z D, where g is a function analytic on D. Proof. By the power series representation in (3) the result is true, at least in, with n=1 (4) g(z) = a 1 + a 2 (z z 0 ) + a 3 (z z 0 ) 2 + (z ). In other words, if we define g(z) = f(z)/(z z 0 ) for z D\{z 0 } and g(z 0 ) = a 1, then in g is given by (4), and so is analytic in. But in D\{0} it s the quotient of two analytic functions, with the 1 That s just the way the things are... don t make to much of it! 1

2 2 denominator never vanishing, hence it s also analytic on D\{0}, hence on all of D. 2. Order We continue with the standing assumption that f is a function that is analytic on a domain D of the plane. Now let s assume it s not identically zero on D, but that some z 0 D is a zero of f. We want to examine the structure this imposes on f. Although we didn t mention this in the last section, the coefficient a 1 in (3) may also be zero. Since we assume f isn t identically zero in D, it follows from (1) some that coefficient isn t zero. Let a m be the first non-zero coefficient, so (5) a 0 = a 1 = = a m 1 = 0 but a m 0. In this case we say z 0 is a zero of order m. For z 0 a zero of order m the power series representation for f in is: (6) f(z) = a m (z z 0 ) m + a m+1 (z z 0 ) m+1 + = (z z 0 ) m g(z) where g(z) = a m +a m+1 (z z 0 )+a m+2 (z z 0 ) 2, the series converging in. In particular, g is analytic in and g(z 0 ) = a m 0. With this definition we can refine Theorem 1.1 as follows: 2.1. Theorem. Suppose f(z 0 ) = 0. Then the following three statements are equivalent: (a) z 0 is a zero of f of order m. (b) f(z 0 ) = f (z 0 ) = f (m 1) (z 0 ) = 0 but f (m) (z 0 ) 0. (c) f(z) = (z z 0 ) m g(z) for z D, where g is analytic in and g(z 0 ) 0. Proof. (a) (b): This follows from the definition of order m and formula (2) giving power series coefficients in terms of derivatives. (a) (c): We observed this just before stating the theorem. (c) (a): Given (c), we have a power series expansion for g, valid in ; write it as: g(z) = a m + a m+1 (z z 0 ) + a m+2 (z z 0 ) 2, where g(z 0 ) = a m 0. Then multiply both sides of the above equation by (z z 0 ) m to get the power series expansion of f, which you now see begins with a m (z z 0 ) m, where a m 0. Thus f has a zero of order m

3 at z 0. The proof that g is actually analytic in all of D follows just as it did in the proof of Theorem 1.1, and I leave the argument to you Exercise. Show that: If f has a zero of order m at z 0. Then f n (n-th power of f) has a zero of order mn there Exercise. Show that: If f has a zero of order m at z 0, and h is analytic in D with h(z 0 ) 0, then fh has a zero of order m at z Exercise. More generally, show that: If f has a zero of order m at z 0 and h has a zero of order n there, then fh has a zero of order mn at z Example. The zeros of sin z all have order 1. Proof I. We saw on a previous handout that the zeros of sin z are just the ones we learned in Calculus: z 0 = nπ, where n runs through all integers. The way of seeing that nπ has order one is to differentiate: If f(z) = sin z then f (z) = cos z, so f (nπ) = cos nπ = ±1 0, hence the zero nπ has order 1 by part (b) of the above Theorem. Proof II. We know that sin z has power series expansion: sin z = z z3 3! + z5 5!, valid for all z C. This shows right away that 0 is a zero of order 1 in the sine function. In particular the expansion above is still valid if z nπ replaces z. But sin(z nπ) = ± sin z (with the plus sign if n is even and the minus sign if n is odd). Thus ] (z nπ)3 (z nπ)5 sin z = ± [(z nπ) +, 3! 5! valid for all z C, so by the definition of order of a zero, nπ has order Exercise. Show that each zero of sin 2 z has order 2. (Hint: use the result of a previous exercise.) 2.7. Contrast with Calculus I.. We ve seen above that if f is analytic on a domain D and not identically zero there, then each zero of f has integer strength (i.e. it has some positive integer order). This quantum phenomenon for zeros of analytic functions has no analogue for differentiable functions of a real variable! For example, consider the function f(x) = x 5/3, defined for all real x. It s differentiable at every point of R, but its lone zero, at the origin, does not have integer order 3

4 4 f(x) (in the sense that lim x 0 does not exist (finitely) for any positive x n integer n). If anything, we should say the origin is a zero of order 5/ Exercise. (a) Write down some more examples of this kind, where differentiable functions of a real variable can have fractional or even irrational orders. (b) Let f(z) = z 5/2. Why doesn t f serve as a counterexample to the statement that: zeros of analytic functions have integer orders? (c) Look ahead to the example of Remark 3.6 (if you don t wnat to look ahead, it s f(x) = x 2 forx 0 and f(x) = 0 if x < 0). What order would you assign to the zero x 0 = 1/2? 3. Uniqueness Theorems. It s clear from the power series representation (1), and the expression (2) of the series coefficients in terms of derivatives, that if all the derivatives of f at z 0 are zero, then f is identically zero in. In fact more is true: 3.1. Theorem. If f (n) (z 0 ) = 0 for each n = 0, 1, 2,..., then f 0 on D. Proof. (cf. pp of our text). Note that if D = C, i.e., if f is entire, then this is easy, for then we can take the disc to be the whole complex plane, so the result follows from the comment that led off this section. Suppose D is not the whole plane. Fix z in D. We wish to show that f(z) = 0. To do this, join z 0 to z by a polygonal curve Γ in D. 2 Write 0 for the original disc (the largest one centered at z 0 and contained in D, and draw a finite succession of open discs 1,... n, each with center on Γ, each having its center in the previous disc, and with n centered at z. 3 We ve already noted that f vanishes identically on 1. Thus it, and all its derivatives, vanish at the center of 2, hence (same argument with 2 replacing 1 ) it vanishes identically on 2. Repeat the argument, eventually arriving at the fact that f vanishes identically on n, so in particular at its center, z. 2 This is possible because D is connected. The textbook s definition of connectedness is: every pair of points in D can be joined by a polygonal curve lying entirely in D 3 Draw a picture, at least for Γ a straight line, to convince yourself that this can be done.

5 3.2. Corollary: The First Uniqueness Theorem. Suppose f and g are analytic on a domain D, and for some point z 0 D, f (n) (z 0 ) = g (n) (z 0 ) for n = 0, 1, 2,.... Then f g on D. 5 Proof. Apply Theorem 3.1 to the analytic function f g Corollary. The Second Uniqueness Theorem. Suppose: f and g are analytic on a domain D, {z 1, z 2,...} is a sequence in D that converges to a point z 0 of D, and f(z n ) = g(z n ) for each n = 1, 2,.... Then f g on D. Proof. Suppose first that g 0 on D, i.e. that f 0 on the set {z n } 1. Our goal is to show that f 0 on D. Since f is continuous on D, f(z 0 ) = lim n f(z n ) = 0, hence f f(z n ) f(z 0 ) (z 0 ) = lim = 0. 4 n z n z 0 Thus (by the work of 2) we have f(z) = (z z 0 )g(z) where g is analytic on D, and since f vanishes at each point z n, so does g. Using the result just proved, with g in place of f, we see that g (z 0 ) = 0. By the Product Rule: f (z) = (z z 0 )g (z) + g(z) so f (z) = (z z 0 )g (z) + 2g (z), hence f (z 0 ) = 0. Continuing in this fashion we see that f and all its derivatives vanish at z 0, hence, by Theorem 3.1, f 0 on D. To obtain the result for general g, just apply the result of the last paragraph to f g Remark. The Second Uniqueness Theorem shows that our extensions of the the trigonometric functions and the exponential function from the real line to analytic functions on the plane are unique. For example, suppose you ve found a function that s analytic on a domain D that contains a segment of the real line, and you know that your function takes the value sin x for each x in that segment. Then by the Second Uniqueness Theorem, you know your function must coincide with sin z on D. 5 4 because the numerator of the difference quotient is zero for each n. 5 because that segment contains lots of sequence that converge to points of D... right?

6 Exercise. Suppose you ve defined an function f analytic on D = C\{(, 0]} that agrees with the natural logarithm on the positive real axis. Show that f is the principal branch of the logarithm Remarks. (a) Nothing like these uniqueness theorems for analytic functions can be true for differentiable functions of a real variable. For example, the function f : R R defined by { x 2 if x 0 f(x) = 0 if x < 0 is differentiable at each point of R, identically zero on the negative real axis, but not identically zero on the real line. (b) Example (a) can be refined further: There exist functions f : R R that are infinitely differentiable on R and vanish identically on the negative real axis, but are not identically zero on R. One such example is: { e 1/x if x > 0 f(x) = 0 if x Exercise. (20 HW points!! Due last day of class). Show that for function f of part (b) above: f has derivatives of all orders at each point of the real line, 6 and that f (n) (0) = 0 for each n = 0, 1, 2,.... In particular, note that f has a Taylor series, center at the origin, that converges to zero for all x R. Lesson: Even if a Taylor series converges in some real interval, it may not converge to the function it s supposed to represent! 7 6 The origin is the only point in question here. 7 Remember: this never happens for analytic functions!