14.3 Cauchy’s integral formula 14.4 Derivatives of analytic functions
Eugenia Malinnikova, NTNU
October 24 2016
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Simply connected domains and Cauchy’s integral theorem
A domain D on the complex plain is said to be simply connected if any simple closed curve in D is a boundary of a subdomain of D. Example 1. Any circle is a simply connected domain. 2. A circular ring or a punched disc are not simply connected domains. Theorem Let f be an analytic function in a simply connected domain D. If C is a simple closed curve in D then I f (z)dz = 0 C
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Integration of analytic functions along paths
Corollary R In a simply connected domain the integral C f (z)dz of an analytic function does not depend on the path C but only on its end points, we write also
Z Z ze f (z)dz = f (z)dz C z0
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Anti-derivative
Theorem Let f be an analytic function on a simply connected domain D. Then there is an analytic function F in D such that F 0(z) = f (z) for each z in D and Z f (z)dz = F (ze ) − F (z0) C
where C is a simple curve with end points z0 and ze .
To construct the anti-derivative we fix some point zc in D and for each z in D define Z z F (z) = f (ζ)dζ zc One can check that F defined in this way is analytic and F 0(z) = f (z).
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Cauchy’s integral formula
Theorem Let f be an analytic function in a simply connected domain D. If C is a simple closed curve in D that encloses a point z0 then I f (z) dz = 2πif (z0) C z − z0
To prove the formula we write f (z) = f (z0) + (f (z) − f (z0)) and f (z) f (z ) f (z) − f (z ) = 0 + 0 z − z0 z − z0 z − z0 The proof for Cauchy’s integral theorem implies that I f (z) I f (z) dz = dz C z − z0 K z − z0
where K is a small circle centered at z0. Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Proof of Cauchy’sintegral formula
After replacing the integral over C with one over K we obtain Z Z 1 f (z) − f (z0) f (z0) dz + dz K z − z0 K z − z0 The first integral is equal to 2πi and does not depend on the radius of the circle. The second one converges to zero when the radius goes to zero (by the ML-inequality).
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Example
Assume that C encloses z0 then H z I dz = 2πiz C z−z0 0 H ez z I dz = 2πie 0 C z−z0 H z3 H z3 1 H z3 H z3 I C z2+1 dz = C (z+i)(z−i) dz = 2i C z−i dz − C z+i dz If C encloses both i and −i then we apply the Cauchy’s formula to both integrals
I 3 z 1 3 3 2 dz = (2πii − 2πi(−i) ) = −2πi C z + 1 2i
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Derivatives of analytic functions
Theorem If f is analytic in some domain D then it has derivatives of any order which are also analytic functions. Moreover,
1 I f (z) f (z0) = dz, 2πi C z − z0 I 0 1 f (z) f (z0) = 2 dz, 2πi C (z − z0) I (n) n! f (z) f (z0) = n+1 dz, 2πi C (z − z0) where C is a simple closed path in D that bounds some domain in D which contains z0.
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Examples
I I z 2 dz = 2πi C (z − z0) I I z z0 e 2πie z 000 z 4 dz = , (e ) = e C (z − z0) 6 I I cos z −2πi cos π πi 00 3 dz = = , (cos z) = − cos z C (z − π) 24 12
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Cauchy’s inequality
Suppose that f is an analytic function in a disc of radius r around z0 and that |f (z)| ≤ M when |z − z0| = r. Then n!M |f (z)| ≤ M, and |f (n)(z )| ≤ , for |z − z | ≤ r. 0 r n 0
Let C = {z : |z − z0| = r}, we have I (n) n! f (z) f (z0) = n+1 dz 2πi C (z − z0) then taking the absolute values and applying ML-inequality, we get n! M n!M |f (n)(z )| ≤ 2πr = 0 2π r n+1 r n
Eugenia Malinnikova, NTNU TMA4120, Lecture 19 Applications
Theorem (Liouville) If a function is analytic in a whole complex plane and bounded in absolute value, then it is a constant Functions analytic in the whole plane are called entire functions. Theorem (Morera) If f is continuous in a simply connected domain D and I f (z)dz = 0 C for any closed curve C. Then f is analytic in D.
Eugenia Malinnikova, NTNU TMA4120, Lecture 19