MATH 417 Homework 2 Instructor: D. Cabrera Due June 23. v = e x sin y

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1 MATH 47 Homework Instructor: D. Cabrera Due June under the trans-. Find and sketch the image of the rectangle 0 < x <, π < < π formation w = e z. Solution: The transformation w = e z can be written as w = e z = e x cos + ie x sin If we define w = u + iv then the functions u and v are u = e x cos, v = e x sin Let s consider the four boundaries: i. The top of the rectangle is = π, 0 x. In this case, the functions u and v are u = e x cos π = 0, v = ex sin π = ex Thus, the image is the vertical line u = 0 where e 0 v e or v e. ii. The right side of the rectangle is x =, π π. In this case, the functions u and v are u = e cos = e cos, v = e sin = e sin Thus, the image is the right half of the circle u + v = e. iii. The bottom of the rectangle is = π, 0 x. In this case, the functions u and v are ( u = e x cos π ) ( = 0, v = e x sin π ) = e x Thus, the image is the vertical line u = 0 where e v e 0 or v. iv. The left side of the rectangle is x = 0, π π. In this case, the functions u and v are u = e 0 cos = cos, v = e 0 sin = sin Thus, the image is the right half of the circle u + v =. Since the sides of the rectangle are not included in the set, their images are not included.

2 4 v u Sketch the following sets and determine whether the are open, closed, or neither. (a) z + < (b) Im z < (c) 0 < z < (d) Rez = (e) z 4 z Solution: (a) The set z + < is the set of points inside the circle of radius centered at z =. The set is open because it does not include an of its boundar points, which are all points on the circle z+ = x

3 (b) The set Imz < is the set of points satisfing < or < <. This is a vertical strip between the lines = and =. The set is open because it does not include an of its boundar points, which are all points on the lines = and = x (c) The set 0 < z < is the set of points inside the circle of radius centered at z = except for the point z =. The set is open because it does not include an of its boundar points, which are all points on the circle z = and the point z = x x (d) The set Re z = is the set of points satisfing x =. The set is closed because it contains all of its boundar points, which are all points in the set. - - The set z 4 z can be better described b simplifing the inequalit as follows: (e) z 4 z (x 4) + i x + i (x 4) + x + (x 4) + x + x 8x x + 8x x The set is open because it includes all of its boundar points, which are the points on the line x =. - 4 x - -

4 . Show that (a) lim z z = (b) lim z z + z 4 = Solution: (a) To show that the limit is, we use the fact that lim f(z) = lim z z 0 z z0 f(z) = 0. Therefore, since we know that lim z z 0 f(z) = lim z z z = lim = 0 z lim f(z) = lim z z 0 z z =. (b) To show that the limit is, we use the fact that Therefore, since lim f(z) = w 0 lim f z z 0 lim f z 0 ( ) ( z = lim ) + z z 0 ( z ) 4 = lim z 0 ( ) = w 0. z + z 4z = we know that lim f(z) = lim z z z + z 4 =. 4. Find f (z) for the following functions: (a) f(z) = 4z + 5z (b) f(z) = ( z ) (c) f(z) = z + z where z Solution: Using the derivative rules, we have (a) f (z) = 8z + 5 (b) f (z) = 6z ( z )

5 (c) f (z ) (z + ) (z) = = (z ) 8 (z ) 5. Show that f (z) does not exist at an point z when f(z) = Rez. Solution: First, we have f(z) = Rez = Re (x+i) = x. So u(x, ) = x and v(x, ) = 0. These functions have continuous first derivatives everwhere in the complex plane. The first partial derivatives are u x =, v = 0 u = 0, v x = 0 Clearl, the first Cauch-Riemann equation u x = v is never satisfied as 0. Therefore, the function f(z) is not differentiable anwhere. 6. Let z = x + i. Determine the values of z for which the Cauch-Riemann equations are satisfied for the following functions: (a) f(z) = e x e i (b) f(z) = x + ix (c) f(z) = x + i (d) f(z) = Im z Solution: (a) First we rewrite the given function as f(z) = e x e i = e x (cos( ) + i sin( )) = e x cos ie x sin Therefore, u = e x cos and v = e x sin. The first partial derivatives are u x = e x cos, u = e x sin, v = e x cos v x = e x sin We can see that the Cauch-Riemann equations (u x = v, u = v x ) are satisfied for all z = x + i. (b) Here we have u = x and v = x. The first partial derivatives are u x =, v = x u = 0, v x = In order for the second of the C-R equations to be satisfied, we must have u = v x 0 = = 0

6 However, if = 0 then the first of the C-R equations gives us u x = v = x = x(0) = 0 which is impossible. Therefore, the C-R equations are never satisfied. (c) Here we have u = x and v =. The first partial derivatives are u x = x, v = u = 0, v x = 0 The second of the C-R equations, u = v x, is alwas satisfied as 0 = 0. In order for the first of the C-R equations to be satisfied we must have u x = v x = x = Therefore, the C-R equations are satisfied for all z satisfing z = x + ix where x is an real number. (d) First we rewrite f(z) as f(z) = Im z = Im (x + i) = Therefore we have u = and v = 0. The first partial derivatives are u x = 0, v = 0 u =, v x = 0 The second of the C-R equations is never satisfied because u = v x = 0 is impossible. Therefore, the C-R equations are never satisfied.