Now we compute the derivative. d dz log z = e −ıθ ∂ ∂r (Log r + ıθ) = e −ıθ 1 r = 1 z Solution 8.14 The complex derivative in the coordinate directions is d dz = e −ıθ ∂ ∂r = − ı r e −ıθ ∂ ∂θ . We substitute f = u + ıv into this identity to obtain the Cauchy-Riemann equation in polar co ordinates. e −ıθ ∂f ∂r = − ı r e −ıθ ∂f ∂θ ∂f ∂r = − ı r ∂f ∂θ u r + ıv r = − ı r (u θ + ıv θ ) We equate the real and imaginary parts. u r = 1 r v θ , v r = − 1 r u θ u r = 1 r v θ , u θ = −rv r Solution 8.15 Since w is analytic, u and v satisfy the Cauchy-Riemann equations, u x = v y and u y = −v x . 414 Using the chain rule we can write the derivatives with respect to x and y in terms of u and v. ∂ ∂x = u x ∂ ∂u + v x ∂ ∂v ∂ ∂y = u y ∂ ∂u + v y ∂ ∂v Now we examine φ x − ıφ y . φ x − ıφ y = u x Φ u + v x Φ v − ı (u y Φ u + v y Φ v ) φ x − ıφ y = (u x − ıu y ) Φ u + (v x − ıv y ) Φ v φ x − ıφ y = (u x − ıu y ) Φ u − ı (v y + ıv x ) Φ v We use the Cauchy-Riemann equations to write u y and v y in terms of u x and v x . φ x − ıφ y = (u x + ıv x ) Φ u − ı (u x + ıv x ) Φ v Recall that w  = u x + ıv x = v y − ıu y . φ x − ıφ y = dw dz (Φ u − ıΦ v ) Thus we see that, ∂Φ ∂u − ı ∂Φ ∂v =  dw dz  −1  ∂φ ∂x − ı ∂φ ∂y  . We write this in operator notation. ∂ ∂u − ı ∂ ∂v =  dw dz  −1  ∂ ∂x − ı ∂ ∂y  415 The complex conjugate of this relation is ∂ ∂u + ı ∂ ∂v =  dw dz  −1  ∂ ∂x + ı ∂ ∂y  Now we apply both these operators to Φ = φ.  ∂ ∂u + ı ∂ ∂v  ∂ ∂u − ı ∂ ∂v  Φ =  dw dz  −1  ∂ ∂x + ı ∂ ∂y  dw dz  −1  ∂ ∂x − ı ∂ ∂y  φ  ∂ 2 ∂u 2 + ı ∂ 2 ∂u∂v − ı ∂ 2 ∂v∂u + ∂ 2 ∂v 2  Φ =  dw dz  −1   ∂ ∂x + ı ∂ ∂y  dw dz  −1   ∂ ∂x − ı ∂ ∂y  +  dw dz  −1  ∂ ∂x + ı ∂ ∂y  ∂ ∂x − ı ∂ ∂y   φ (w  ) −1 is an analytic function. Recall that for analytic functions f, f  = f x = −ıf y . So that f x + ıf y = 0. ∂ 2 Φ ∂u 2 + ∂ 2 Φ ∂v 2 =  dw dz  −1   dw dz  −1  ∂ 2 ∂x 2 + ∂ 2 ∂y 2   φ ∂ 2 Φ ∂u 2 + ∂ 2 Φ ∂v 2 =     dw dz     −2  ∂ 2 φ ∂x 2 + ∂ 2 φ ∂y 2  Solution 8.16 1. We consider f(z) = log |z| + ı arg(z) = log r + ıθ. The Cauchy-Riemann equations in polar coordinates are u r = 1 r v θ , u θ = −rv r . 416 We calculate the derivatives. u r = 1 r , 1 r v θ = 1 r u θ = 0, −rv r = 0 Since the Cauchy-Ri emann equations are satisfied and the partial derivatives are continuous, f(z) is analytic in |z| > 0, |arg(z)| < π. The complex derivative in terms of polar coordinates is d dz = e −ıθ ∂ ∂r = − ı r e −ıθ ∂ ∂θ . We use this to differentiate f(z). df dz = e −ıθ ∂ ∂r [log r + ıθ] = e −ıθ 1 r = 1 z 2. Next we consider f(z) =  |z| e ı arg(z)/2 = √ r e ıθ/2 . The Cauchy-Riemann equations for polar coordinates and the polar form f(z) = R(r, θ) e ıΘ(r,θ) are R r = R r Θ θ , 1 r R θ = −RΘ r . We calculate the derivatives for R = √ r, Θ = θ/2. R r = 1 2 √ r , R r Θ θ = 1 2 √ r 1 r R θ = 0, −RΘ r = 0 Since the Cauchy-Ri emann equations are satisfied and the partial derivatives are continuous, f(z) is analytic in |z| > 0, |arg(z)| < π. The complex derivative in terms of polar coordinates is d dz = e −ıθ ∂ ∂r = − ı r e −ıθ ∂ ∂θ . 417 We use this to differentiate f(z). df dz = e −ıθ ∂ ∂r [ √ r e ıθ/2 ] = 1 2 e ıθ/2 √ r = 1 2 √ z Solution 8.17 1. We consider the function u = x Log r −y arctan(x, y) = r cos θ Log r − rθ sin θ We compute the Laplacian. ∆u = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 1 r ∂ ∂r (cos θ(r + r Log r) − θ sin θ) + 1 r 2 (r(θ sin θ − 2 cos θ) − r cos θ Log r) = 1 r (2 cos θ + cos θ Log r − θ sin θ) + 1 r (θ sin θ − 2 cos θ − cos θ Log r) = 0 The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations. v r = − 1 r u θ , v θ = ru r v r = sin θ(1 + Log r) + θ cos θ, v θ = r (cos θ(1 + Log r) −θ sin θ) We integrate the first equation with respect to r to determine v to within the constant of integration g(θ). v = r(sin θ Log r + θ cos θ) + g(θ) We differentiate this expression with respect to θ. v θ = r (cos θ(1 + Log r) −θ sin θ) + g  (θ) 418 We compare this to the second Cauchy-Riemann equation to see that g  (θ) = 0. Thus g(θ) = c. We have determined the harmonic conjugate. v = r(sin θ Log r + θ cos θ) + c The corresponding analytic function is f(z) = r cos θ Log r − rθ sin θ + ı(r sin θ Log r + rθ cos θ + c). On the positive real axis, (θ = 0), the function has the value f(z = r) = r Log r + ıc. We use analytic continuation to determine the function in the complex plane. f(z) = z log z + ıc 2. We consider the function u = Arg(z) = θ. We compute the Laplacian. ∆u = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 0 The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations. v r = − 1 r u θ , v θ = ru r v r = − 1 r , v θ = 0 We integrate the first equation with respect to r to determine v to within the constant of integration g(θ). v = −Log r + g(θ) 419 We differentiate this expression with respect to θ. v θ = g  (θ) We compare this to the second Cauchy-Riemann equation to see that g  (θ) = 0. Thus g(θ) = c. We have determined the harmonic conjugate. v = −Log r + c The corresponding analytic function is f(z) = θ − ı Log r + ıc On the positive real axis, (θ = 0), the function has the value f(z = r) = −ı Log r + ıc We use analytic continuation to determine the function in the complex plane. f(z) = −ı log z + ıc 3. We consider the function u = r n cos(nθ) We compute the Laplacian. ∆u = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 1 r ∂ ∂r (nr n cos(nθ)) − n 2 r n−2 cos(nθ) = n 2 r n−2 cos(nθ) − n 2 r n−2 cos(nθ) = 0 420 The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations. v r = − 1 r u θ , v θ = ru r v r = nr n−1 sin(nθ), v θ = nr n cos(nθ) We integrate the first equation with respect to r to determine v to within the constant of integration g(θ). v = r n sin(nθ) + g(θ) We differentiate this expression with respect to θ. v θ = nr n cos(nθ) + g  (θ) We compare this to the second Cauchy-Riemann equation to see that g  (θ) = 0. Thus g(θ) = c. We have determined the harmonic conjugate. v = r n sin(nθ) + c The corresponding analytic function is f(z) = r n cos(nθ) + ır n sin(nθ) + ıc On the positive real axis, (θ = 0), the function has the value f(z = r) = r n + ıc We use analytic continuation to determine the function in the complex plane. f(z) = z n 4. We consider the function u = y r 2 = sin θ r 421 We compute the Laplacian. ∆u = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 1 r ∂ ∂r  − sin θ r  − sin θ r 3 = sin θ r 3 − sin θ r 3 = 0 The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations. v r = − 1 r u θ , v θ = ru r v r = − cos θ r 2 , v θ = − sin θ r We integrate the first equation with respect to r to determine v to within the constant of integration g(θ). v = cos θ r + g(θ) We differentiate this expression with respect to θ. v θ = − sin θ r + g  (θ) We compare this to the second Cauchy-Riemann equation to see that g  (θ) = 0. Thus g(θ) = c. We have determined the harmonic conjugate. v = cos θ r + c The corresponding analytic function is f(z) = sin θ r + ı cos θ r + ıc 422 [...]... ıvx f (z) = 2 e x2 −y 2 (x cos(2xy) − y sin(2xy)) + 2 ex f (z) = 2 e x2 −y 2 2 −y 2 (y cos(2xy) + x sin(2xy)) ((x + ıy) cos(2xy) + (−y + ıx) sin(2xy)) Finding the derivative is easier if we first write f (z) in terms of the complex variable z and use complex differentiation 2 −y 2 f (z) = ex (cos(2x, y) + ı sin(2xy)) f (z) = ex 2 −y 2 eı2xy f (z) = e(x+ıy) f (z) = ez 2 f (z) = 2z ez 424 2 2 Solution... neighborhood of any point, it is nowhere analytic 423 2 We calculate the first partial derivatives of u and v u x = 2 ex 2 −y 2 uy = 2 e (x cos(2xy) − y sin(2xy)) x2 −y 2 (y cos(2xy) + x sin(2xy)) vx = 2 e x2 −y 2 (y cos(2xy) + x sin(2xy)) vy = 2 e x2 −y 2 (x cos(2xy) − y sin(2xy)) Since the Cauchy-Riemann equations, ux = vy and uy = −vx , are satisfied everywhere and the partial derivatives are continuous, f (z)... dimensions and let {ξi } be an orthogonal coordinate system The distance metric coefficients hi are defined 2 ∂x1 ∂ξi hi = ∂x2 ∂ξi + 2 The Laplacian is 2 u= ∂ ∂ξ1 1 h1 h2 h2 ∂u h1 ∂ξ1 + ∂ ∂ 2 h1 ∂u h2 ∂ 2 First we calculate the distance metric coefficients in polar coordinates hr = hθ = 2 ∂x ∂r + 2 ∂x ∂θ + ∂y ∂r ∂y ∂θ 2 = cos2 θ + sin2 θ = 1 2 = r2 sin2 θ + r2 cos2 θ = r Then we find the Laplacian 2 φ= 1... arg(z + 1) The velocity potential and a branch of the stream function are plotted in Figure 8.10 The stream lines, arg(z − 1) + arg(z + 1) = c, are plotted in Figure 8.11 Next we find the velocity field v= 2 v= 2 φ 2x(x + y − 1) 2y(x2 + y 2 + 1) ˆ ˆ x+ 4 y x4 + 2x2 (y 2 − 1) + (y 2 + 1 )2 x + 2x2 (y 2 − 1) + (y 2 + 1 )2 431 Figure 8.9: Velocity field and velocity direction field for φ = ln r − θ The velocity... domains D1 and D2 , respectively Suppose that D1 ∩ D2 is a region or an arc and that f1 (z) = f2 (z) for all z ∈ D1 ∩ D2 (See Figure 9.4.) Then the function f (z) = f1 (z) for z ∈ D1 , f2 (z) for z ∈ D2 , is analytic in D1 ∪ D2 D1 D2 D1 D2 Figure 9.4: Domains that Intersect in a Region or an Arc Result 9.1 .2 follows directly from Result 9.1.1 9 .2 Analytic Continuation of Sums Example 9 .2. 1 Consider... + ıc z Solution 8.18 1 We calculate the first partial derivatives of u = (x − y )2 and v = 2( x + y) ux uy vx vy = 2( x − y) = 2( y − x) =2 =2 We substitute these expressions into the Cauchy-Riemann equations ux = vy , uy = −vx 2( x − y) = 2, 2( y − x) = 2 x − y = 1, y − x = −1 y =x−1 Since the Cauchy-Riemann equation are satisfied along the line y = x−1 and the partial derivatives are continuous, the function... (z) and f2 (z) be analytic functions defined in D If f1 (z) = f2 (z) for the points in a region or on an arc in D, then f1 (z) = f2 (z) for all points in D To prove Result 9.1.1, we define the analytic function g(z) = f1 (z) − f2 (z) Since g(z) vanishes in the region or on the arc, then g(z) = 0 and hence f1 (z) = f2 (z) for all points in D 439 Result 9.1 .2 Consider analytic functions f1 (z) and f2 (z)... into the complex plane An obvious choice for f (z) is f (z) = cos z sin z Using trig identities we can write this as f (z) = sin(2z) 2 Example 9.3.7 Find f (z) given only that u(x, y) = cos x cosh2 y sin x + cos x sin x sinh2 y 445 Recall that f (z) = ux + ıvx = ux − ıuy Differentiating u(x, y), ux = cos2 x cosh2 y − cosh2 y sin2 x + cos2 x sinh2 y − sin2 x sinh2 y uy = 4 cos x cosh y sin x sinh y f... = 1 ∂ r ∂r r ∂φ ∂r + 1 2 r2 ∂ 2 We calculate the partial derivatives of u ∂u ∂r ∂u r ∂r ∂ ∂u r ∂r ∂r 1 ∂ ∂u r r ∂r ∂r ∂u ∂θ ∂2u ∂ 2 1 ∂2u r2 ∂ 2 = cos θ + log r cos θ − θ sin θ = r cos θ + r log r cos θ − rθ sin θ = 2 cos θ + log r cos θ − θ sin θ = 1 (2 cos θ + log r cos θ − θ sin θ) r = −r (θ cos θ + sin θ + log r sin θ) = r ( 2 cos θ − log r cos θ + θ sin θ) = 1 ( 2 cos θ − log r cos θ + θ sin... the two forms 2 We verify that log z is analytic for r > 0 and −π < θ < π using the polar form of the Cauchy-Riemann equations Log z = ln r + ıθ 1 1 ur = v θ , uθ = −vr r r 1 1 1 = 1, 0 = −0 r r r 426 Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous for r > 0, log z is analytic there We calculate the value of the derivative using the polar differentiation formulas . analytic. 423 2. We calculate the first partial derivatives of u and v. u x = 2 e x 2 −y 2 (x cos(2xy) − y sin(2xy)) u y = 2 e x 2 −y 2 (y cos(2xy) + x sin(2xy)) v x = 2 e x 2 −y 2 (y cos(2xy) + x sin(2xy)) v y =. ∇φ v = 2x(x 2 + y 2 − 1) x 4 + 2x 2 (y 2 − 1) + (y 2 + 1) 2 ˆ x + 2y(x 2 + y 2 + 1) x 4 + 2x 2 (y 2 − 1) + (y 2 + 1) 2 ˆ y 431 Figure 8.9: Velocity field and velocity direction field for φ = ln r −θ. The. x direction. f  (z) = u x + ıv x f  (z) = 2 e x 2 −y 2 (x cos(2xy) − y sin(2xy)) + 2 e x 2 −y 2 (y cos(2xy) + x sin(2xy)) f  (z) = 2 e x 2 −y 2 ((x + ıy) cos(2xy) + (−y + ıx) sin(2xy)) Finding the derivative