COMPLEX
Bread and Butter Topics (1) Contour Integrals (2) Rouche's Theorem and counting zeros (3) Conformal Maps
Other Minor Topics (which havent been covered) (1) Analytic Continuation / Re ection Principle (2) Classi cation of Singularities (3) Schwazrz Christoel maps (4) Harmonic Functions/ Poisson Integral Formula
1. Weierstrass Product Formula Lemma 1. Q∞ converges P∞ converges P∞ converges n=N (1 + an) ⇔ n=N log(1 + an) ⇔ n=N |an| Proof. The rst two are related by taking a log one way and an exp the other way. The second two are related by
log(1+an) the limit comparison test limn→∞ = 1 . an Example 2. Q∞ z converges. This has zeros at 2. f(z) = j=1 1 − j2 zj = j
Proof. By the lemma, we just need to show that P |z| . This converges uniformly on compact subsets, because j2 < ∞ for we have P |z| P 1 |z| < R j2 < R j2 < ∞ Example 3. Q∞ z does NOT converge. f(z) = j=1 1 − j
Proof. We need P |z| . But since P 1 , this does not converge! j < ∞ j ≮ ∞
∞ z Example 4. Q z j converges. f(z) = j=1 1 − j e
h z z i z z z z z2 z2 z z Proof. Notice that log 1 − e j = +log 1 − = − − ... ≤ C . So then for |z| ≤ R Q 1 − e j ≤ j j j j j j2 j2 j 2 C z 2 Q j2 Q R which converges just as in the rst example! |e | ≤ 1 + 2C j2
2 3 n Proposition 5. Let z z z . For a given sequence with , choose En(z) = (1 − z) exp z + 2 + 3 + ... + n zj |zj| → ∞ n +1 P R j any sequence nj so that < ∞ for any R > 0. Then the function: zj ∞ Y z f(z) = E nj z j=1 j
converges and has zeros at z = zj.
Proof. The proof is very similar to the examples above. The function En(z) is chosen exactly so that in the taylor n+1 series expansion for log(En(z)) the rst n terms cancel, and we remain with O z . The condition on nj then ensures that we have convergence of the function.
Remark 6. Notice that the choice nj = j will make the condition we want happen, because if |zj| → ∞ then R 1 P 1 < eventually, so by comparing to the sum j < ∞ we have convergence. In practice however, you can zj 2 2 choose much smaller values that work, see the examples! 1 COMPLEX 2
Theorem 7. (Weierstrass Product Formula) Suppose f(z) is an entire function with zeros at z1, z2,... and a zero of order m at z = 0 (m = 0 is legit too). Then there is a sequence nj and an entire function g(z) so that: ∞ Y z f(z) = zmeg(z) E nj z j=1 j Q∞ z f(z) m Q∞ z Proof. By the above proposition, we nd nj so that En makes sense. But then /z En j=1 j zj j=1 j zj f(z) m Q∞ z g(z) has no zeros! By the lemma that follows, /z En = e for some entire g(z), which completes the j=1 j zj proof. Lemma 8. If h(z) is entire and has no zeros, then h(z) = eg(z)for some entire function g(z).
Proof. One way to see the proof is to draw a picture of the range of h, namely h(C) this is simply connected and does not contain zero, so we can de ne a branch of the logarithm here. Composing the functions gives a g so that g(z). Another way to see this is to see that h0 is an entire function, so it has an antiderivative, say 0 h0 . h(z) = e h g = h −g 0 0 0 −h −g But then (e h) = (−g h + h ) e = 0, so this must be constant and then e = h up to a constant factor. Theorem 9. (Hadamard Theorem) If f(z) is like in the statement of the Weierstrass Product Formula, and we α have the additional property that |f(z)| ≤ Cc|z| , then the function g(z) in the conclusion of the Weierstrass product formula is a polynomial with degree ≤ bαc.
Proof. HARD! 2. More Conformal Maps Problem 10. Map the region between and 1 1 onto the upper half plane. |z| = 1 |z − 2 | = 2 Problem 11. Map the region C\{z : |z| = 1, Im z ≥ 0} to the outside of a unit circle, and so that the point at in nity in the domain gets mapped to the point at in nity in the range. Problem 12. Describe the image of the unit disk under the map 1 . z 7→ log 1−z 3. More Contour Integrals Problem 13. Verify P∞ 1 π2 by integrating 1 around the boundary of a suitable box. Detailed 1 n2 = 6 z2(e2πiz −1) estimates please. Problem 14. ∞ cos mx = ? 0 1+x4 dx ´ 4. More Contour Integrals Problem 15. How many zeros does the polynomial z6 + 3z4 + 1 have in 1 ≤ |z| ≤ 2? In |z| ≤ 1? Characterize the relative positions of the roots as completely as possible.