University of British Columbia Math 301 Midterm 2 March 16, :00-11:50am

Size: px

Start display at page:

Download "University of British Columbia Math 301 Midterm 2 March 16, :00-11:50am"

Randolf Miller
5 years ago
Views:

1 University of British Columbia Math 301 Midterm 2 March 16, :00-11:50am Last Name (print): First Name (print): Student ID Number: Signature: Instructor: Richard Froese Instructions: 1. No notes, books or calculators are allowed. 2. Read the questions carefully and make sure you provide all the information that is asked for in the question. 3. Show all your work. Answers without any explanation or without the correct accompanying work could receive no credit, even if they are correct. 4. Answer the questions in the space provided. Continue on the back of the page if necessary. Run L A TEX again to produce the table

2 y 2e i3π/4 φ = 1 2e iπ/4 φ = 0 φ = Figure 1: The region R x 1. Let R be the region in the upper half plane lying above the circle centred at 0 with radius 2 (see Figure 1.). (a) (5 points) Find a conformal map f(z) that maps R onto the upper half plane. (Hint: use the Joukowski map J(z) = (z + 1/z)/2 together with another map.) Find the values of f(2e iπ/4 ) and f(2e i3π/4 ). Solution: Perform as scaling z z/2 followed by the Joukowski map. This results in f(z) = J(z/2) = (z/2+2/z)/2. We have f(2e iπ/4 ) = (e iπ/4 +e iπ/4 )/2 = cos(π/4) = 1/ 2 and f(2e i3π/4 ) = (e i3π/4 + e i3π/4 )/2 = cos(3π/4) = 1/ 2 (b) (5 points) Solve Laplace s equation φ(x, y) = 0 for x + iy in R with boundary conditions φ = 0 on the part of the boundary of R lying on the real axis as well as on the portion of the circle {2e iθ : 0 θ π/4, 3π/4 θ π}, and φ = 1 on the remaining part of the boundary {2e iθ : π/4 θ 3π/4}. Solution: We must first solve the problem on the image of R under the conformal map f. This is a Dirichlet problem on the upper half plane with boundary conditions Φ(w) = 0 for w = u real and u < 1/ 2 or u > 1/ 2 and Φ(w) = 1 for w = u real and 1/ 2 < u < 1/ 2. The solution is a linear combination A Arg(w+1/ 2)+B Arg(w 1/ 2)+C.This satisfies the boundary conditions if A = 1/π, B + 1/π and C = 0. Thus Φ(w) = (1/π) Arg(w + 1/ 2) + (1/π) Arg(w 1/ 2) The solution φ(x, y) is then given by φ(x, y) = Φ(f(x + iy)) Page 2

3 v w Arg(w + 1/ 2) Arg(w 1/ 2) 1/ 2 1/ 2 u Figure 2: The image of R and the dotted line under f (c) (5 points) The function φ(x, y) is discontinuous at the boundary point 2e iθ/4. Nevertheless, if we approach 2e iθ/4 along the dotted line, the corresponding values of φ will converge. Compute this limit. (Hint: The angle between the dotted line and the tangent line to the boundary circle is preserved under a conformal map.) Solution: The angle between the dotted line and the tangent line to the boundary circle is π/4. The conformal property then tells us that the image of the dotted line under f will meet the point 1/ 2 at an angle of π/4. So as w 1/ 2 along the image of the dotted line, Arg(w 1/ 2) π/4. Also Arg(w + 1/ 2) 0 since w is approaching a point to the right of 1/ 2 on the boundary. Thus Φ(w) 1/4 and this is also the limiting value of φ that we seek. Page 3

4 2. (a) (5 points) Find a fractional linear transformation f(z) such that f(i) = 0, f(0) = 1 + i and f( i) = 2 Solution: We first find the FLT that maps i, 0, i to 0, 1,. This is the cross ratio (z, i, 0, i) = z + i z + i [ ] 1 i corresponding to the matrix or any non-zero multiple of this. Next we 1 i find the FLT that maps 0, 1 + i, 2. This is the cross ratio ( 1 + i)w (w, 0, 1 + i, 2) = (1 + i)w 2(1 + i) [ ] 1 + i 1 + i corresponding to the matrix or any non-zero multiple of this i The map we want is the first one followed by the inverse of the second. This is associated to the matrix [ ] [ ] [ ] 2 2i 1 i 1 i 2 2i 2 2i = i 1 i 2 2 This yields (dropping a factor of 2) f(z) = (1 i)z 1 i z 1 Page 4

5 (b) (5 points) What ares the images of the imaginary axis and of the real axis under f? Solution: Since three points on the imaginary axis get mapped to three points on the circle {z : z 1 = 1} and the map is an FLT, the image of the whole real axis must be the circle {z : z 1 = 1}. The real axis makes an angle of π/2 with the imaginary axis at z = 0, so by the conformal property the image of the real axis under f must make an angle of π/2 with the circle {z : z 1 = 1} at f(0) = 1 + i. Moreover, the image of the real axis contains f(1) =. This implies that the image of the real axis is the vertical line through 1 + i. Alternatively we could pick three points on the real axis, say 0, 1,, and compute their image under f. This yields 1 + i,, 1 i which all lie on the the vertical line through 1 + i. (c) (5 points) Describe how you could find the point α, such that α and 1/2 are symmetric with respect to the circle { z 1 = 1}. You do not need to carry out the calculations. Solution: Since the circle { z 1 = 1} is the image of the imaginary axis, the points 1/2 and α are symmetric with respect to the circle { z 1 = 1} if the points f 1 (1/2) and f 1 (1/2) are symmetric with respect to the imaginary axis. This means that they must be related by a reflection across this axis. [ ] i So we could start with 1/2, then compute f 1 (1/2) using the matrix. 1 1 i (This yields 3/5 + 4i/5.) Now reflect accross the imaginary axis. (This yields 3/5 + 4i/5.) Then α = f(3/5 + 4i/5). ( 1). You could also recall the formula for the symmetric point, or that the symmetric point to z lies on the line connecting z with the origin of the circle at a distance of R 2 /r from the centre (where R is the radius of the circle and r the distance from z to the centre. In our case R = 1, r = 1/2 so R 2 1/r = 2. The centre of the circle is at 1 so that α = 1. Page 5

6 3. (10 points) How many zeros does p(z) = z 6 + z have in the right half plane? Solution: We must compute the change in argument as z travels down the imaginary axis. To do this set z = iy for y real. Then p(iy) = y 6 +iy 5 +1 so Re(p(iy)) = y 6 +1 and Im(p(iy)) = y 5. The real values of y where either Re(p(iy)) or Im(p(iy)) vanish are 1, 0, 1. We can now track the behaviour of the argument of p(iy) as follows: y p(iy) aligns with the negative real axis y = 1 Re(p) = 0, Im(p) > 0 so p(iy) aligns with the positive imaginary axis y = 0 Re(p) > 0, Im(p) = 0 so p(iy) aligns with the positive real axis y = 1 y Re(p) = 0, Im(p) < 0 so p(iy) aligns with the negative imaginary axis p(iy) aligns with the negative real axis So the argument changes by 2π as z travels down the imaginary axis. The total change of argument going around a D shaped contour is 6π 2π = 4π which yields N + (p) = 4π/(2π) = 2 as the number of zeros in the right half plane. y y = 1 y y y = 0 x y = 1 Page 6

The University of British Columbia. Mathematics 300 Final Examination. Thursday, April 13, Instructor: Reichstein

The University of British Columbia. Mathematics 300 Final Examination. Thursday, April 13, 2017. Instructor: Reichstein Name: Student number: Signature: Rules governing examinations Each examination candidate