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Complex Analysis Study Guide Complex Analysis Study Guide 1. Equivalences Of Holomorphicity For a domain D ⊂ C, f(z) = u + iv is holomorphic in D if and @f only if = 0: @z (a) If and only if u and v satisfy the Cauchy-Riemann Equations Proof: @f 1 @ @ 1 @u @v i @v @u @u @v @u @v 0 = = + i (u+iv) = − + + () = and = − : @z 2 @x @y 2 @x @y 2 @x @y @x @y @y @x (b) If and only if f(z) 2 C1(D). Proof: First assume f 0 2 C1(D). f(z) − f(z ) f(z + h) − f(z ) u(x + h; y ) + iv(x + h; y ) − u(x ; y ) − iv(x ; y ) lim 0 = lim 0 0 = lim 0 0 0 0 0 0 0 0 z!z0 z − z0 h!0;h2R h h!0 h u(x0 + h; y0) − u(x0; y0) v(x0 + h; y0) − v(x0; y0) @u @v = lim + i lim = + i : h!0 h h!0 h @x @x (x0;y0) (x0;y0) On the other hand f(z) − f(z ) f(z + ih) − f(z ) lim 0 = lim 0 0 z!z0 z − z0 h!0;h2R ih 1 u(x0; y0 + h) − u(x0; y0) v(x0; y0 + h) − v(x0; y0) @u @v = lim + i lim = (−i) + : h!0 i h h!0 h @y @y (x0;y0) (x0;y0) 0 Equating the real and imaginary parts for f (z0), we get @u @v @u @v = ; = i @x @y @y @x evaluated at (x0; y0). By part (a), f is holomorphic. Now assume f is holomorphic on D. Fix z0 2 D. For z near z0 we may define γ(t) = (1 − t)z0 + tz and γ : [0; 1] ! D. I @f Z 1 @f dγ Z 1 @f f(z) − f(z0) = f(γ(1)) − f(γ(0)) = dz = (γ(t)) dt = (γ(t))(z − z0)dt: γ @z 0 @z dt 0 @z Dividing both sides by z − z0 yields f(z) − f(z ) Z 1 @f 0 = (γ(t))dt z − z0 0 @z Z 1 @f Z 1 @f @f @f Z 1 @f @f = (z0)dt + (γ(t)) − (z0) dt = (z0) + (γ(t)) − (z0) dt: 0 @z 0 @z @z @z 0 @z @z @f Let > 0. As @z is continuous, there is a δ > 0 such that @f @f (w) − (z0) < @z @z whenever jw − z0j < δ. Note that jγ(t) − z0j = tjz − z0j ≤ jz − z0j for t 2 [0; 1], therefore @f @f (γ(t)) − (z0) < @z @z 1 for jz − z0j < δ. Fixing jz − z0j < δ, we have Z 1 Z 1 Z 1 @f @f @f @f (γ(t)) − (z0) dt ≤ (γ(t)) − (z0) dt ≤ dt = . 0 @z @z 0 @z @z 0 It follows that f(z) − f(z0) @f lim = (z0): z!z0 z − z0 @z 2. Interaction of Holomorphic Functions (a) O(D), the set of all holomorphic functions on D forms an algebra. Proof: Let f; g; h 2 O(D) and a; b 2 C. To be an algebra, the following conditions must hold: • (f + g) · h = f · h + g · h. • f · (g + h) = f · g + f · g. • (af) · (bg) = (ab)(f · g). As the algebra multiplication is standard commutative multiplication, it suffices to show that if f and g are any holomorphic function on D, then fg is as well. @(fg) @f @g = g + f = 0g + f0 = 0: @z @z @z So O(D) is in fact an algebra. (b) Moreover, if f : D ! E is holomorphic and g : E ! C is holomorphic, then g ◦ f is holomorphic. Proof: As f is holomorphic, f(x; y) = uf (x; y) + vf (x; y) and g(x; y) = ug(x; y) + vg(x; y). u(x; y) + iv(x; y) = g ◦ f = g(uf ; vf ) = ug(uf ; vf ) + ivg(uf ; vf ): @u @ @ug @uf @ug @vf = (ug(uf ; vf )) = + @x @x @uf @x @vf @x and @v @v @u @v @v = g f + g f : @y @uf @y @vf @y As g is holomorphic, @u @v @u @v g = g ; g = − g @uf @vf @vf @uf and as f is holomorphic @u @v @u @v f = f ; f = − f @x @y @y @x So @u @v @u @v = ; = − ; @x @y @y @x so g ◦ f is holomorphic. (c) If u is harmonic on a domain E, f : D ! E is holomorphic, then u ◦ f is harmonic in D. Proof: Let z0 2 E. As E is a domain, there exists a r > 0 such that D(z0; r) ⊂ E. As D(z0; r) is simply connected, on this set we can find some holomorphic g such that Re(g) = u. On D(z0; r), g ◦ f is holomorphic, so its real part is harmonic. Notice Re(g ◦ f) = u ◦ f. 3. Harmonic Conjugates and Antiderivatives 2 (a) If f = u + iv is holomorphic, then u and v are harmonic. v is called the harmonic conjugate of u. Proof: @u @v @u @v By the Cauchy-Riemann Equations, = and = − : Thus @x @y @y @x @2u @2u @ @u @ @u @ @v @ @v + = + = + − : @x2 @y2 @x @x @y @y @x @y @y @x As f 2 C1(D), order of differentiation can be changed, so @2u @2u @v @v + = − = 0: @x2 @y2 @x@y @x@y The computation to show v is harmonic is identical. @f @g Lemma 1. If f; g are C1 functions on a simply connected set D and if = on D, then there is @y @x a function h 2 C2(D) such that @h @h = f; = g @x @y on D. Proof: For (x; y) 2 D, set Z x Z y h(x; y) = f(t; b)dt + g(x; s)ds; a b where (a; b) 2 D. As D is simply connected, these integrals exist. By the Fundamental Theorem of Calculus @h (x; y) = g(x; y): @y Again by the Fundamental Theorem of Calculus and since g 2 C1(D), @h @ Z y Z y @ Z y @ (x; y) = f(x; b) + g(x; s)ds = f(x; b) + g(x; s)ds = f(x; b) + f(x; s)ds @x @x b b @x b @y = f(x; b) + f(x; y) − f(x; b) = f(x; y): (b) Given a harmonic function u on a domain D, if D is simply connected u has a harmonic conjugate v such that F = u + iv is holomorphic in D. Give a counterexample if D is not simply connected. Proof: @u @u @f @g Let f = − , g = . As u is harmonic, we have = on D. Since f; g 2 C1(D), by @y @x @y @x Lemma 1, there exists a v 2 C2(D) such that @v @u @v @u = f = − ; = g = : @x @y @y @x By the Cauchy-Riemann Equations, F = u + iv is holomorphic. Let D be the punctured unit disc. u(x; y) = ln(x2 + y2) is harmonic on D. Moreover, u along with the function v(x; y) = 2 tan−1(y=x) satisfies the Cauchy-Riemann equations. However, v is not defined everywhere on D, so u + iv is not holomorphic on D. (c) If f is holomorphic in a domain D, prove that if D is simply connected then there exists a holomorphic function F such that F 0(z) = f(z) on D. Give a counterexample if D is not simply connected. Proof: 3 @g @h Let f = u + iv. Let g = u, h = −v, then by the Cauchy-Riemann equations we have = . @y @x 2 By Lemma 1 there exists a real C function f1 such that @f @f 1 = g = u; 1 = h = −v: @x @y ^ @g^ @h^ Now letg ^ = v; h = u. Again by the Cauchy-Riemann equations, we get @y = @x so by the Lemma there exists a function f2 such that @f @f 2 =g ^ = v; 2 = h^ = u: @x @y 2 Let F = f1 + if2. F is C and by the above equations F satisfies the Cauchy-Riemann equations, so F is holomorphic. Finally, @ 1 @ @ 1 @f @f i @f @f 1 i F = − i (f +f ) = 1 + 2 + 2 − 1 = (u+u)+ (v +v) = f: @z 2 @x @y 1 2 2 @x @y 2 @x @y 2 2 Let D be the punctured unit disc. Let f(z) = 1=z. f is holomorphic on D. Assume there exists a I holomorphic function F such that F 0(z) = f(z). Let γ = @D(0; 1). By calculation f(z) = 2πi. γ I However, by the Fundamental Theorem of Calculus, f(z) = F (γ(1)) − F (γ(0)) = 0. So no F γ can exist. 4. Cauchy's Theorem, Morera's Theorem and Cauchy's Integral Formula (a) Cauchy's Theorem D is a bounded domain in C with piecewise C1 boundary. If f is holomorphic in D and f 2 C(D), then I f(z)dz = 0: @D Proof: By Stokes' Theorem, I Z Z @f @f Z @f f(z)dz = d(f(z)dz) = dz + dz ^ dz = dz ^ dz + 0dz ^ dz = 0: @D D D @z @z D @z (b) Morera's Theorem Let D be connected. If f 2 C(D) and for any simply closed piecewise C1 curve γ in D we have I f(z)dz = 0, then f is holomorphic in D. γ Proof: 1 Fix z0 2 D. Define a function F : D ! C as follows. Given z 2 D choose a piecewise C curve φ : [0; 1] ! C such that φ(0) = z0, φ(1) = z. Set Z F (z) = f(w)dw: φ 1 It is not yet clear that F is well-defined.
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