3.1. COMPLEX DERIVATIVES 89 Theorem 3.1.4 (Chain rule) Let f : D D′ C and g : D′ C be holomorphic → ⊂ → functions on the domains D and D′ respectively. Suppose that both f and g are C1-smooth.10 Then (g f)′(z)= g′(f(z)) f ′(z), z D. ◦ · ∀ ∈ Proof: z D, write w = f(z ). By the C1-smooth condition and Taylor Theorem, we ∀ 0 ∈ 0 0 have f(z + h)= f(z )+ f ′(z )h + o(h), and g(w + h) g(w )= g′(w )h + o(h), 0 0 0 0 − 0 0 so that ′ g(f(z0 + h)) g(f(z0)) = g(w0 + hf (z0)+ o(h)) g(w0) ′ ′− ′ − = g (w0) hf (z0)+ o(h) + o hf (z0)+ o(h) ′ ′ = g (w0)f (z0)h + o(h). Then g(f(z + h)) g(f(z )) g′(w )f ′(z )h + o(h) 0 − 0 = 0 0 g′(w )f ′(z )= g′(f(z ))f ′(z ). h h → 0 0 0 0 [Example] Let f(z)= sin(z2 3z + 1). Then f ′(z)= cos(z2 +3z + 1) (2z 3) by the chain rule. − · − Analytic Function Must be Holomorphic So far we have defined holomorphic func- tions. What are examples of holomorphic functions ? The following theorem says that any analytic function must be holomorphic. In fact, later we shall see the converse is also true, namely, any holomorphic function must be analytic. ∞ n Theorem 3.1.5 Let f(z) = an(z a) be convergenet, a C, with the radius of n=0 − ∈ convergence R> 0. Then f isP a holomorphic function on ∆(a, R) and ∞ ′ n−1 f (z)= nan(z a) , z ∆(a, R). − ∀ ∈ Xn=1 10This C1-smooth condition will be dropped later by Goursat theorem. 90 CHAPTER 3. HOLOMORPHIC FUNCTIONS ∞ n−1 Proof:(Optional reading) Claim 1: g(z) := nan(z a) has the radius of conver- n=1 − gence R. P To prove, we assume a = 0. Then by Theorem 2.4.2, 1 1 n−1 1 1 n n n n ln an n n n = lim a = elim n ln |a | = e lim n−1 | | lim n = elim n−1 ln|a | R n→∞ n p| | n n 1 = lim a = lim (n + 1) a = n→∞ n+1 n→∞ n+1 R′ p| | p | | Here in the last second equality, we used the fact that √n n +1 1 as n . Claim 1 is proved. It remains to show: → →∞ f ′(z)= g(z), z ∆(0, R). (3.3) ∀ ∈ In fact, f(z + h) f(z) − g(z) ?? 0, as h 0 h − → → ∞ n n (z + h) z n−1 = an − nz h − Xn=1 ∞ n n n−v−1 v = an z h v Xn=2 Xv=2 ∞ n n n−v−1 v an z h ≤ | | v | | | | Xn=2 Xv=2 ∞ n n ( z + h ) z n−1 = an | | | | −| | n z | | h − | | Xn=1 | | ∞ n n n−1 n ( z + h ) z n z = an r | | | | −| | | | (we choose r with z + h <r<R) | | h rn − rn | | | | Xn=1 | | ∞ 1 z + h n z n n z n−1 M | | | | | | | | ≤ h r − r − r r Xn=1 | | 1 z + h z r = M | | | | | | (we use geometric series) h r ( z + h ) − r z − (r z )2 | | − | | | | −| | −| | Mr h = | | (r z )2(r ( z + h )) −| | − | | | | 0, as h 0, → → 3.1. COMPLEX DERIVATIVES 91 n where M > 0 is a constant such that anr M, n. (3.3) is proved. | | ≤ ∀ Corollary 3.1.6 f is analytic = f is holomorphic. ⇒ Remark: 1. We’ll prove later, by Cauchy’s Integral Formula, that f is holomorphic = f is analytic. (3.4) ⇒ 2. By the theory of power series, an analytic function has derivatives of any order, while by the definition of holomorphic function, a holomorphic function is only required to have the differentiation of first order. These two definitions are so different, but they are equivalent. 3. While Cauchy studied complex-valued functions f with f ′ exists, Weierstrass published an extensive assay in 1876 of his “ systematic foundation of analysis” of analytic functions. This very influential paper dealt with the problem of representation of single-valued complex functions. Jacobian of holomorphic functions Let f(z)= u(x, y)+ iy(x, y) be a complex-valued function. We can regard f as a map from an open subset of R2 to R2, (x, y) (u, v), so that its Jacobian is defined by 7→ ∂u ∂u ∂x ∂y Jf(z) := det ∂v ∂v ∂x ∂y where z = x + iy. The following result shows that the Jacobian of a holomorphic function is always non negative. Proposition 3.1.7 Let f be a holomorphic function. Then Jf(z)= f ′(z) 2. | | 92 CHAPTER 3. HOLOMORPHIC FUNCTIONS Proof: ∂u ∂v ∂u ∂v ∂u 2 ∂u 2 Jf(z)= = + = f ′(z) 2, ∂x ∂y − ∂y ∂x ∂x ∂y | | ′ ∂u ∂v where we used the Cauchy-Riemann Equations and the formula f (z)= ∂x + i ∂x . Preserving Angle Property Let C be a non-singular curve given in parametric form φ :(α, β) C, t x(t)+ iy(t). → 7→ Such curve can be understood as a vector-valued function φ :(α, β) R2, t (x(t),y(t)) if we identify R2 with C. Suppose that C is non singular (i.e., if φ′(z→)= x′(t)+7→iy′(t) exists and = 0). For6 each t, if we regard φ(t)= x(t)+ iy(t) as a vector (x(t),y(t)), φ′(t)= x′(t)+ iy′(t) can be regarded as the tangent vector (x′(t),y′(t)) of the curve C at the point φ(t). The curve C at the point φ(t) has a tangent vector whose direction is determined by arg(φ′(t)). Now suppose that C is in a domain D C. Let f : D C be a holomorphic function. Then we obtain the image curve f(C) given⊂ by the parametric→ function ψ := f φ :(α, β) C, t f(φ(t)). ◦ → 7→ By the chain rule (Theorem 3.1.5), ψ′(t)= f ′(φ(t))φ′(t). Then arg(ψ′(t)) = arg(f ′(z)) + arg(φ′(t)), mod(2π). (3.5) Therefore, the tangent vector of the curve f(C) at f(φ(t)) is changed from the tangent vector of the curve C at φ(t), and such change is caused by arg(f ′(φ(t))), which is called the angle of rotation. [Example] Let f(z) = cos z. Determine the angle of rotation given by arg(f ′(z)) at the points z1 = i, z2 = 1 and z3 = π + i. Proof: f ′(z)= sinz so that − e−1 e f ′(z )= f ′(i)= sin i = − = i sinh 1, 1 − − 2i − 3.1. COMPLEX DERIVATIVES 93 f ′(z )= f ′(1) = sin 1, 2 − ei(π+i) e−i(π+i) f ′(z ) = f ′(π + i)= sin(π + i)= − 3 − − 2i eiπ e−1 e−iπ e ( 1)e−1 ( 1)e e e−1 = · − · = − − − = i − = i sinh 1. 2i − 2i 2 eiz−e−iz ex−e−x Here sin z = 2i and sinh x = 2 as in calculus. Therefore, the angle of rotation is given by π π α = arg( i sinh 1) = , α = arg( sin 1) = π, α = arg(i sinh 1) = . 1 − − 2 2 − 3 2.