Math 417 Midterm Exam Solutions Friday, July 9, 2010

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1 Math 417 Midterm Exam Solutions Friday, July 9, 010 Solve any 4 of Problems 1 6 and 1 of Problems 7 8. Write your solutions in the booklet provided. If you attempt more than 5 problems, you must clearly indicate which problems should be graded. Answers without justification will receive little to no credit. 1. a) Evaluate 1 + i) 50 and write your answer in the form a + bi. b) Find all values of log i). c) Find all solutions to the equation z = 8. Write your answers in the form a+bi. a) The modulus of z = 1 + i is z = and the principal argument is Θ = π 4. Using DeMoivre s Theorem we have 1 + i) 50 = r 50 cos 50Θ + i sin 50Θ) ) i) 50 = cos 150π + i sin 150π ) i) 50 = 5 cos 75π ) 75π + i sin We note that the angle 75π is equivalent to π 75π since 18π) = π. Therefore, 1 + i) 50 = 5 cos π + i sin π ) = 5 i b) The modulus of z = i is z = and the principal argument is Θ = π. Using the definition of log z we have where k = 0, ±1, ±,.... log z = ln r + i Θ + kπ) log i) = ln + i π ) + kπ c) The solutions to the equation are the cube roots of 8. We use the formula: ) )] Θ + kπ Θ + kπ z 1/ = r cos 1/ + i sin, k = 0, 1,

2 The modulus of z = 8 is z = r = 8 and the principal argument is Θ = π. Therefore, the solutions are ) )] π + 0)π π + 0)π z 1/ = 8 cos 1/ + i sin = cos π + i sin π ) = 1 + i ) )] π + 1)π π + 1)π z 1/ = 8 cos 1/ + i sin = cosπ + i sin π) = ) )] π + )π π + )π z 1/ = 8 cos 1/ + i sin = cos 5π + i sin 5π ) = 1 i. a) Sketch the set of points defined by the inequality z + z. b) Sketch the image of the set of points in the z-plane defined by π x π, 0 y < under the transformation w = sin z. a) Letting z = x + iy we have z + z x + iy + x + iy x + ) + iy x + iy x + ) + y x + y x + ) + y x + y x + 4x y x + y 4x x 1 Part a) y Part b) v x u

3 b) First, we want to write sin z in terms of x and y. w = sinz) = sin x cosh y + i cos x sinh y We consider w to be complex and write w = u i v so that u = sin x cosh y v = cosxsinh y Now let s consider the transformation of each boundary. i. For π x π, y = 0 we have u = sin x cosh 0 = sin x v = cosxsinh 0 = 0 Thus, the transformation is the line segment 1 u 1, v = 0. ii. For x = π, 0 y < we have u = sin v = cos π ) cosh y = cosh y π ) sinh y = 0 Thus, the transformation is the line segment < u 1, v = 0. iii. For x = π, 0 y < we have u = sin π cosh y = cosh y v = cos π sinh y = 0 Thus, the transformation is the line segment 1 u <, v = 0. Putting these together, the boundary of the given region is transformed into the entire u-axis in the w-plane. Now take a test point in the given region, say, z = 0 + i. Then we have u = sin 0 cosh 1 = 0 v = cos 0 sinh 1 = e e 1 > 0 ) e e 1 The transformation of z = 0+i is w = 0 + i which is in the upper half of the w-plane.

4 . In each part below, z is a complex number. a) Show that z = z for all z. b) Find the values of z, if any, for which e z = e z. a) This can be proven in one of two ways. If we let z = x + iy then z = x + iy) z = x y ) + ixy) z = x y ) + xy) z = x 4 x y + y 4 + 4x + y z = x 4 + x y + y 4 z = x + y ) z = x + y z = z for all z. If, instead, we let z = re iθ then z = r e iθ = r e iθ = r = z for all z. b) If we let z = x + iy then e z = e x+iy e z = e x e iy e z = e x cosy + i sin y) e z = e x cosy i sin y) e z = e x e iy e z = e x iy e z = e z for all z. 4. onsider the function fz) = z z. a) Write the function in the form fz) = ux, y) + ivx, y). b) Find all values of z for which f z) exists.

5 a) Let z = x + iy. Then z = x iy and we have fz) = z z fz) = x + iy) x iy) fz) = x y ) + ixy) ] x iy) fz) = xx y ) + xy + i yx y ) + x y ] fz) = x + xy + iy + x y) b) Let ux, y) = x + xy and vx, y) = y + x y. Both ux, y) and vx, y) have continuous derivatives of all orders everywhere in the complex plane. The first partial derivatives are u x = x + y, v y = y + x u y = xy, v x = xy In order for the auchy-riemann equations to be satisfied we need u x = v y x + y = y + x u y = v x xy = xy x = y 4xy = 0 x = ±y xy = 0 The second equation says that either x = 0 or y = 0. If x = 0 then the first equation says that = 0. If y = 0 then the first equation says that x = 0. Thus, the -R equations are only satisfied when z = 0 and f z) exists only when z = Determine the values of z for which the function fz) = ze x is differentiable and evaluate f z) at each point. At what points, if any, is fz) analytic? Let z = x + iy. Then fz) = ze x fz) = x iy)e x fz) = xe x + i ye x ) We have ux, y) = xe x and vx, y) = ye x. These functions have continous derivatives of all orders everywhere in the complex plane. The first partial derivatives are u x = xe x + e x, v y = e x u y = 0, v x = ye x

6 In order for the auchy-riemann equations to be satisfied we need u x = v y xe x + e x = e x u y = v x 0 = ye x xe x + e x = 0 ye x = 0 e x x + ) = 0 Since e x > 0 for all x, the first equation tells us that x = and the second equation tells us that y = 0. Therefore, the -R equations are only satisfied when z = and f z) exists only when z =. There is no neighborhood of z = throughout which f z) exists. Thus, fz) is analytic nowhere. 6. onsider the function ux, y) = xy x y + x 6y. a) Show that ux, y) is harmonic in the entire complex plane. b) Find a harmonic conjugate vx, y) of ux, y). a) The function ux, y) has continuous derivatives of all orders everywhere in the complex plane. The first and second partial derivatives are u x = y x y +, u xx = 6xy u y = xy x 6, u yy = 6xy We can see that u xx + u yy = 6xy + 6xy = 0 for all x, y. Therefore, ux, y) is harmonic in the entire complex plane. b) A harmonic conjugate vx, y) of ux, y) must satisfy the auchy-riemann equations. v y = u x v x = u y v y = y x y + v x = xy + x + 6 Integrating the first equation with respect to y we have v y dy = y x y + ) dy vx, y) = 1 4 y4 x y + y + φx)

7 Differentiating this equation with respect to x and setting the result equation to the equation for v x above we get x vx, y) = v x 1 x 4 y4 ) x y + y + φx) = xy + x + 6 xy + φ x) = xy + x + 6 φ x) = x + 6 φx) = x + 6) dx φx) = 1 4 x4 + 6x + Therefore, the family of harmonic conjugates of ux, y) are vx, y) = 1 4 x4 + y 4 ) x y + y + 6x + 7. onsider the integral I = z z ) dz where is the circle z = oriented counterclockwise. a) Use the ML-Bound formula to find an upper bound on I. b) Find the exact value of I. a) First, the length of the contour is L = πr = π) = 4π. Next, we find an upper bound on fz) for all z on using the Triangle Inequality. z z z + z = z + z = + = 6 Therefore, we let M = 6 and we get the following upper bound on I : I ML = 64π) = 4π b) The function fz) = z z is analytic nowhere. So we have to parametrize the

8 contour. Let zt) = e it where 0 t π. Then z = e it, z t) = ie it, and fz) dz = z z) dz = b a π fzt))z t) dt 0 π 4e it e it) ie it ) dt = 4i e it 1 ) dt 0 = 4i ] π i e it t 0 = 4i i e πi π + ] i e0 + 0 = 4i i π + ] i = 8πi So the exact value of I is I = 8π. 8. Evaluate the integral where the contour is z dz a) the line segment with initial point 1 and final point i b) the arc of the unit circle z = 1 traversed in the clockwise direction with initial point 1 and final point i. Why don t the two results agree? a) A parametrization of is zt) = t+it+1) where 1 t 0. Then z t) = 1+i

9 and we have fz) dz = z dz = b a 0 1 = 1 + i) fzt))z t) dt t + t + 1) ] 1 + i) dt 0 1 t + t + 1) dt ] 0 = 1 + i) t + t + t 1 = 1 + i) 0 )] = 1 + i) b) A parametrization of is zt) = e it where π t π. Then z t) = ie it and fz) dz = z dz = = b a π/ π π/ π fzt))z t) dt e it ie it ) dt ie it dt = e it π/ π = e iπ/ e iπ = i + 1 The results do not agree because fz) is analytic nowhere so the integral is not path independent.