LECTURE-15 : LOGARITHMS AND COMPLEX POWERS VED V. DATAR∗ The purpose of this lecture is twofold - first, to characterize domains on which a holomorphic logarithm can be defined, and second, to show that the only obstruction to defining a holomorphic logarithm is in defining a continuous logarithm. Throughout this lecture we use the notation, ∗ C = C n f0g: To set the stage, let us revisit the difficulties we had in defining a holo- morphic logarithm. For z = reiθ, consider the function (0.1) log z = log jzj + iθ; where, for instance, we can let θ 2 [θ0; θ0 + 2π). In the first figure below, as we traverse the curve back to the point p, the argument goes from θ0 to θ0 + 2π, and hence the log z does not return to the original value. In other words log z as defined is not continuous. On the other hand, if we return to p along the curve traversed in the second figure, the argument does return to θ0, and log z does not jump in value. The difference between the situations if of course that the first curve goes around 0 while the second does not. Thus logarithm is an example of a multivalued function, and zero in this case is called a branch point. ∗ In general, we can consider any holomorphic function f :Ω ! C . Then, a holomorphic function g :Ω ! C is called a branch of the logarithm of f, and denoted by log f(z), if eg(z) = f(z) for all z 2 Ω. A natural question to ask is the following. ∗ Question 0.1. Given a holomorphic function f :Ω ! C , when can we define a holomorphic branch of log f(z). From the point of view of the Cauchy theory, the multivalued behaviour of the logarithm function is esentially because 1=z, which would be the derivative of a holomorphic logarithm function, does not integrate out to zero around curves that contain the origin in the interior. Keeping this in mind, we have the following basic theorem. ∗ Theorem 0.1. Let Ω be a connected domain, and f :Ω ! C be holomor- phic such that f 0(z) is continuous and such that Z f 0(z) dz = 0 γ f(z) Date: 24 August 2016. 1 for all closed loops γ ⊂ Ω. • Then there exists a holomorphic function g :Ω ! C, denoted by g(z) = log f(z), such that eg(z) = f(z): • g0 = f 0=f, and hence for any fixed p 2 Ω, Z z f 0(w) g(z) = g(p) + dw: p f(w) Remark 0.1. It is easy to see that if a holomorphic logarithm g(z) = log f(z) exists, then g(z) + 2πin is also a logarithm for all n 2 Z, since these are precisely the periods of the exponential function. Conversely, if g1 and g2 are two logarithms, then they have to differ by a multiple of 2πi. The various logarithm functions are called branches. As a corollary of the theorem, we have the following existence theorem for holomorphic log z. Corollary 0.1. Let Ω be a simply connected domain such that 0 2= Ω. Then there is a holomorphic branch of log z on Ω. Moreover, for any p 2 Ω, Z z 1 log z = log p + dz; p z where we integrate along any path from p to z. Remark 0.2. The proof of the above corollary works for defining a holomor- phic branch of log f(z) as long as f 0, and hence f 0=f, is also holomorphic on Ω. It actually follows from the local theory that we will develop that this is indeed the case for any holomorphic function f(z), and hence we can actually define a holomorphic branch of log f(z) on any simply connected domain. Recall that the branch of log z defined by (0.1) is not even a continuous ∗ function over C . This is not a coincidence. Our next theorem, says that continuity in fact, is the only obstruction to define a holomorphic logarithm. g(z) Theorem 0.2. Let Ω ⊂ C and g :Ω ! C be continuous. If e is holo- morphic, then so is g(z). In other words if f(z) is holomorphic, and we can define a continuous log f(z), then log f(z) is automatically holomorphic. Given a logarithm function, and a w 2 C, one can define a holomorphic complex power, by (0.2) zw = ew log z: Analogous to Theorem 0.2 we have the following theorem. 2 n Theorem 0.3. If g :Ω ! C is continuous such that g(z) is holomorphic for some positive integer n, then g(z) itself is holomorphic. Proofs of Theorems 0.1, 0.2, 0.3, and Corollary 0.1 Proof of Theorem 0.1. Fix a point p 2 Ω. For some a 2 C (to be decided later), define g(z) by Z z f 0(w) g(z) = a + dw; p f(w) where we integrate over any curve joining p and z. By the hypothesis, this is independent of the path chosen. Moreover, by the same argument used before, g(z) is holomorphic with f 0(z) g0(z) = : f(z) Now, consider the function F (z) = e−g(z)f(z). Then F 0(z) = −eg(z)g0(z)f(z) + e−g(z)f 0(z) = 0: Since Ω is connected, this implies that F (z) is a constant. Since g(p) = a, F (p) = e−af(p). If we choose a = log jf(z)j+i arg f(z), then ea = f(p), and hence F (p) = 1. Since f is a constant this would imply that eg(z) = f(z) for all z 2 Ω completing the proof. Proof of Corollary 0.1. We apply the theorem above for f(z) = z. Then not only is f 0(z) = 1 continuous, but f 0=f = 1=z is holomorphic on Ω, since Ω does not contain the origin. But then, since Ω is simply connected, for any closed loop γ ⊂ Ω, Z 1 dz = 0: γ z By the theorem, we then have a holomorphic branch of log z on Ω. Before we prove Theorem 0.2 we need to prove an unconditional result on the local existence of logarithms of zero free functions. ∗ Lemma 0.1. Let f :Ω ! C is a holomorphic function. Then for any a 2 Ω, there exists an δ > 0, and a holomorphic function g(z): Dδ(a) ! C such that Dδ(a) ⊂ Ω and eg(z) = f(z): Proof. Consider the disc D"(f(a)) for any " < jf(a)j. Clearly this does not contain 0 by our choice of ". So the function 1=z is holomorphic on this disc, and since the disc is simply connected, Z 1 dz = 0 γ z 3 for all closed loops γ 2 D"(f(a)). By Theorem 0.1 we can then define a holomorphic branch of log z on this disc. Fix any " < jf(a)j, say " = jf(a)j=2. Then by continuity, there exists an δ < R such that jz − aj < δ =) jf(z) − f(a)j < ": In particular, for all z 2 Dδ(a), f(z) 2 D"(f(a)). For z 2 Dδ(a), we can then define g(z) = log ◦(f(z)) as the composition of the branch of the logarithm chosen above with f(z) when z is restricted to the disc Dδ(a). Proof of Theorem 0.2. Let f(z) = eg(z), which by the hypothesis, is holomorphic, and fix a p 2 Ω. Then by openness and lemma 0.1 above, there exists a disc Dδ(p) ⊂ Ω and a holomorphic function gp(z) on Dδ(p) gp(z) such that e = f(z). Then for all z 2 Dδ(p), g(z) − g (z) p 2 : 2πi Z g(z)−gp(z) Now, since g(z) is continuous, 2πi is a continuous function on Dδ(p) which only takes integer values, and hence has to be a constant. That is, g(z) = gp(z) + 2πin; for some fixed n 2 Z. But then g(z) has to be holomorphic in Dδ(p), since gp(z) is holomorphic, and hence is in particular holomorphic at p. Since p was arbitrary, this completes the proof of the theorem. Proof of Theorem 0.3. We do the proof for n = 2. The general case is also similar. Fix p 2 Ω. Let f(z) = g(z)2, which is holomorphic by assumption. Then like above, by openness and lemma 0.1 above, there exists a disc Dδ(p) ⊂ Ω and a holomorphic function lp(z) on Dδ(p) such lp(z) lp(z)=2 2 that e = f(z). But then gp(z) = e also solves gp(z) = f(z). In other words, g (z)2 p = 1; g(z) and hence g(z) = ±gp(z). for each z 2 Dδ(p). But since g and gp are continuous, clearly either g(z) = gp(z) for all z 2 Dδ(p) or g(z) = −gp(z) for all z 2 Dδ(p). In either case, since gp is holomorphic, this also implies that g(z) is holomorphic in a neighborhood of p. Since p was arbitrary, g(z) is holomorphic on Ω. Examples Somewhat vaguely, for a (multi-valued) function g(z), the point z = a is defined to be a branch point if g(z) is discontinuous while traversing an arbitrarily small circle around the point.