MATH 417 Homework 3 Instructor: D. Cabrera Due June 30. sin θ v x = v r cos θ v θ r. (b) Then use the Cauchy-Riemann equations in polar coordinates

Transcription

1 MATH 417 Homewok 3 Instucto: D. Cabea Due June Let a function f(z) = u + iv be diffeentiable at z 0. (a) Use the Chain Rule and the fomulas x = cosθ and y = to show that u x = u cosθ u θ, v x = v cos θ v θ (b) Then use the Cauchy-Riemann equations in pola coodinates u = v θ, u θ = v and the fact that f (z 0 ) = u x + iv x to show that f (z 0 ) = e iθ (u + iv ) Solution: (a) Using the Chain Rule we have u = u x x + u y y u = u x cosθ + u y (1) and u θ = u x x θ + u y y θ u θ = u x ( ) + u y ( cosθ) (2) By multiplying Equation (1) by cos θ and Equation (2) by and subtacting (2) fom (1) we eliminate u y and get u cos θ u θ = u x cos 2 θ + u x sin 2 θ u cos θ u θ = u x u cos θ u θ = u x (3) which is what we wanted to show. By eplacing u with v we get the othe equation v cosθ v θ = v x (4)

2 (b) We will now use the Cauchy-Riemann equations to eplace the u θ tem in Equation (3) with v and the v θ tem in Equation (4) with u and get u x = u cosθ ( v ) = u cosθ + v v x = v cosθ u = v cosθ u Finally, we plug these expessions fo u x and v x into the deivative f (z) = u x +iv x and simplify to get f (z) in pola coodinates. f (z) = u x + iv x = u cos θ + v + i(v cosθ u ) = u (cosθ i ) + v ( + i cosθ) = u (cosθ i ) + iv (cosθ i ) = u (cos( θ) + i sin( θ)) + iv (cos( θ) + i sin( θ)) = u e iθ + iv e iθ f (z) = e iθ (u + iv ) 2. Show that the function f(z) = e y sin x ie y cos x is entie. Solution: Let u(x, y) = e y sin x and v(x, y) = e y cosx. We can see that u and v have continuous deivatives of all odes eveywhee in the complex plane. Futhemoe, the fist patial deivatives of u and v ae u x = e y cosx, u y = e y sin x, v y = e y cosx v x = e y sin x so we can see that the Cauchy-Riemann equations (u x = v y, u y = v x ) ae satisfied fo all x,y. Theefoe, f (z) exists fo all z in the complex plane and f(z) is entie. 3. Show that the function f(z) = xy + iy is not analytic at any point in the complex plane. Solution: Let u(x, y) = xy and v(x, y) = y. The functions have continuous patial deivatives of all odes eveywhee in the complex plane and the fist patial deivatives ae u x = y, v y = 1 u y = x, v x = 0

3 The Cauchy-Riemann equations (u x = v y, u y = v x ) ae only satisfied when y = 1 and x = 0. Recall that a function is analytic at a point z 0 if it is analytic at evey point in some neighbohood of z 0. Since f (z) exists only at z = i, thee is no neighbohood of z = i which has the popety that f (z) exists at evey point in that neighbohood. Thus, f(z) is analytic nowhee. 4. Let u(x, y) = y x 2 + y 2. (a) Show that u(x, y) is hamonic in the domain D which is the set of all points z in the complex plane excluding z = 0. (b) Find the most geneal hamonic conjugate v of u. Solution: (a) Fist, we note that u(x, y) has continuous fist and second deivatives at evey point in D. Now we must show that u xx + u yy = 0. The fist and second patial deivatives ae u x = (x 2 + y 2 ) 2, u xx = 6x2 y 2y 3 (x 2 + y 2 ) 3 u y = x2 y 2 (x 2 + y 2 ) 2, u yy = 2y3 6x 2 y (x 2 + y 2 ) 3 Clealy, the deivatives u xx and u yy add up to 0. Theefoe, u(x, y) is hamonic in D. (b) A hamonic conjugate v(x, y) of u(x, y) must satisfy the Cauchy-Riemann equations. So we must have v y = u x = (x 2 + y 2 ) 2 v x = u y = y2 x 2 (x 2 + y 2 ) 2 Integating the fist of the above equations with espect to y we have v y = (x 2 + y 2 ) 2 v y dy = (x 2 + y 2 ) dy 2 v(x, y) = (x 2 + y 2 ) dy 2

4 To evaluate the integal we let w = x 2 + y 2, dw = 2y dy. x v(x, y) = w dw 2 v(x, y) = x w + φ(x) v(x, y) = x x 2 + y 2 + φ(x) To find the function φ(x) we diffeentiate the above expession fo v(x, y) with espect to x to get v(x, y) = x x x x 2 + y + 2 x φ(x) v x = y2 x 2 (x 2 + y 2 ) 2 + φ (x) The second of the Cauchy-Riemann equations tells us that v x = y2 x 2 (x 2 + y 2 ) 2 so it must be the case that φ (x) = 0, i.e. φ(x) = C = constant. Theefoe, the most geneal hamonic conjugate of u(x, y) is v(x, y) = x x 2 + y 2 + C 5. Find all values of each expession. (a) exp (2 π ) 4 i (b) log ( 2 + 2i) (c) Log (ei) Solution: (a) exp (2 π ) [ ( 4 i = e 2 cos π ) ( + i sin π )] ( ) 2 2 = e i 2 (b) The modulus and pincipal agument of z = 2 + 2i ae = 2 2, Θ = 3π 4

5 The logaithm of z is then whee k = 0, ±1, ±2,.... log z = ln + i (Θ + 2kπ) ( log( 2 + 2i) = ln 2 ) 2 + i ( 3π 4 + 2kπ (c) The modulus and pincipal agument of z = ei ae The pincipal logaithm of z is then = e, Θ = π 2 Log z = ln + iθ Log (ei) = lne + i π 2 ) Log (ei) = 1 + i π 2 6. Show that the function f(z) = e 2z is entie and wite an expession fo f (z) in tems of z. Solution: It is enough to say that g(z) = 2z and h(z) = e z ae entie so thei composite f(z) = h(g(z)) = e 2z is also entie. Using the Chain Rule, the deivative f (z) is f (z) = 2e 2z We can also solve the poblem by witing f(z) in tems of x and y. f(z) = e 2(x+iy) = e 2x cos 2y + ie 2x sin 2y Note that the functions u(x, y) = e 2x cos 2y and v(x, y) = e 2x sin 2y have continuous fist deivatives eveywhee in the complex plane. Also, the Cauchy-Riemann equations ae satisfied fo all x, y as u x = v y = 2e 2x cos 2y u y = v x = 2e 2x sin 2y Theefoe, f(z) = e 2z is entie. The deivative f (z) is f (z) = u x + iv x f (z) = 2e 2x cos 2y + i(2e 2x sin 2y) f (z) = 2e 2x e i(2y) f (z) = 2e 2(x+iy) f (z) = 2e 2z

6 7. Show that Log ( 1 + i) 2 2 Log ( 1 + i). Solution: Fist, we evaluate the left hand side. The numbe ( 1 + i) 2 can be ewitten as 2i. The modulus and pincipal agument of 2i ae = 2 and Θ = π 2, espectively. Theefoe, the pincipal logaithm of ( 1 + i) 2 is Log ( 1 + i) 2 = ln 2 i π 2 Next, we evaluate the ight hand side. The modulus and pincipal agument of 1 + i ae = 2 and Θ = 3π, espectively. Theefoe, twice the pincipal logaithm of i is ( 2 Log ( 1 + i) = 2 ln 2 + i 3π ) = ln 2 + i 3π 4 2 Clealy, Log( 1 + i) 2 2 Log( 1 + i).