Physics 2400 Midterm I Sample March 2017

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1 Physics 4 Midterm I Sample March 17 Question: Total Points: Gamma function. Leibniz s rule. 1. (1 points) Find positive x that minimizes the value of the following integral I(x) = x+1 x lnγ (u)du. Solution: Using the leibniz s rule let s calculate the derivative of the integral with respect to x: [ ] di Γ (x + 1) = lnγ (x + 1) lnγ (x) = ln. dx Γ (x) Using the relation Γ (x + 1) = xγ (x), the expression can be simplified as following: [ ] di x Γ (x) dx = ln = lnx. Γ (x) The value value of the derivative is at the minimum of I(x). di = when lnx = i.e. x = 1. dx auchy-riemann equations.. (a) (5 points) Out of three functions given below determine (by calculating derivatives only) the one that can be the real part of an analytic function f (z) of a complex argument. 1. u(x,y) = x. u(x,y) = sin(x)cos(y) 3. u(x,y) = x + y (b) (5 points) Use auchy-riemann equations to find the analytic function f (z), z = x + iy from the real part your found above and the additional condition f () =. Page 1 of 7

2 Physics 4 Midterm I Sample March 17 Express the result for f (z) as a function of z only. Solution: auchy-riemann equations, u x = v y and u y = v x, when applied to the function u(x,y) = x + y, produce two first order differential equations: v y = (x + y) = 1, x v x = (x + y) = 1. y Integrating the former relation with respect to y and the latter one with respect to x, obtain v(x,y) = y + c 1 (x) + 1, v(x,y) = x + c (y) +, where c 1 (x) and c (y) are arbitrary real functions of x and y respectively, and 1 and are real integration constants. omparing the last two expression, we conclude that c 1 (x) = x, c (y) = y, 1 = = : The analytic function v(x,y) = x + y +. f (z) = (x + y) + i( x + y) + i = (x + iy) i(x + iy) + i = z (1 i) + i. At z =, Finally, f () = i = =. f (z) = z (1 i). Page of 7

3 Physics 4 Midterm I Sample March 17 omplex algebra. 3. (1 points) Find all solutions of the equation x 3 = 8. Solution: Let s rewrite the right hand side of the equation in the following form: 8 = 3 e in, n =,±1,±,±3,... Taking the cube root from both sides of the equation, arrive at the expression: x 4 = 3 e in n n = e i 3 n, n =,±1,±,±3,... Different values of n label different solutions of the equation. Only three different values of n e.g. n =,1, correspond to different roots. Indeed, Similarly, Therefore, there are three solutions: x 3 = e i = = x. x 4 = x 1, z 5 = z, etc. x n = e i 3 n, n =,1,, or individually z =, z 1 = e i 3, z = e i 4 3. auchy integral theorem. 4. ( points) Show that cosαφ(cosφ) α 1 dφ = α Γ (α) Γ (α), Page 3 of 7

4 Physics 4 Midterm I Sample March 17 (where Γ (x) is Euler s gamma function) by considering the integral J = [z(1 z)] α 1 dz, where the contour (see Fig. 1) a union of the segment of the real axis z = x, x 1, dz = dx, and the semicircular arc z = eiφ, φ, dz = i eiφ dφ. onsider the case α 1. y z = 1 ( 1 + e iφ ) Figure 1: Integration contour for Problem 4. z = x 1 x Solution: Recall the definition of Beta function, 1 B(α,β) = x α 1 (1 x) β 1 dx, and consider the integral: J = [z (1 z)] α 1 dz =, α > 1, Page 4 of 7

5 Physics 4 Midterm I Sample March 17 where the integration is over closed contour shown in Fig. 1. Since the integrand is analytic inside, On the other hand, J =. J = J I + J II, where J I is the integral along the segment of the positive real axis, x 1; J II is the integral along the circular arc or radius R = 1 centered at z = 1. Along the real axis z = x, dz = dx, thus J I = Along the semi-circular arc where < θ <, and Hence, Therefore, J II = ( i where we used that 1 z = eiθ = 1 ei θ ) α ombining, we obtain: x α 1 (1 x) α 1 dx = B(α,α). ( ) e i θ + e i θ = e i θ θ cos, 1 z = 1 1 eiθ = i e i θ sin θ. z(1 z) = i e iθ cos θ sin θ = i eiθ sinθ. dz = i eiθ dθ. e iαθ (sinθ) α 1 dθ = α e i α α e i α i = e i. e iαθ (sinθ) α 1 dθ = B(α,α). e iαθ (sinθ) α 1 dθ, Page 5 of 7

6 Physics 4 Midterm I Sample March 17 Transforming the expression on the left as following: e iα(θ ) (sinθ) α 1 dθ = = e iα(θ ) [ ( cos θ )] α 1 dθ e iαφ [cosφ] α 1 dφ = cos(αφ)[cosφ] α 1 dφ finally obtain the relation: cos(αφ)[cosφ] α 1 dφ = α B(α,α) = α Γ (α) Γ (α). Method of residues. 5. (1 points) Use the method of residues to calculate the following integral. I = e ix x i dx. Sketch your integration contour in the complex plane. Indicate the singularity(s) of the integrand on the same sketch. Solution: Let s consider the following complex integral. J = e iz z i dz where we specify the integration contour later. Page 6 of 7

7 Physics 4 Midterm I Sample March 17 The integrand is exponentially small for Imz +, so we can form a closed contour by complementing the integration path along the real axis with the integral along a large semi-circle in the upper-half z-plane. Per Jordan s lemma, the integral along such a semicircle vanishes. Hence, I = J. The resulting integration contour,, is presented in Fig.. The contour encloses the only singularity of the integrand at z = i: Therefore, ( ) e iz I = J = i Res z i,z = i = i e iz z=i = i e. y Figure : Suitable integration contour in Problem 5. z = Re iθ z = i z = Re iθ R z = x R x Page 7 of 7

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