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Unformatted text preview: Complex Analysis Fall 2007 Homework 5: Solutions 1.5.22 If z = x + iy then z 4 = ( x 4- 6 x 2 y 2 + y 4 ) + i (4 x 3 y- 4 xy 3 ) so that u = Re( z 4 ) = x 4- 6 x 2 y 2 + y 4 and v = Im( z 4 ) = 4 x 3 y- 4 xy 3 . We find that ∂ 2 u ∂x 2 = ∂ ∂x ( 4 x 3- 12 xy 2 ) = 12 x 2- 12 y 2 and ∂ 2 u ∂y 2 = ∂ ∂y (- 12 x 2 y + 4 y 3 ) =- 12 x 2 + 12 y 2 so that ∇ 2 u = ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 = ( 12 x 2- 12 y 2 ) + (- 12 x 2 + 12 y 2 ) = 0 . Since u is a polynomial in x and y its second-order partials are continuous and ∇ 2 u = 0, u is harmonic. We leave the analogous computation involving v to the student. 1.5.32 Claim: If u and v are functions defined on an open set A , u and v satisfy the Cauchy-Riemann equations and u is harmonic on A , then v is a harmonic conjugate of u on A . We are given ∂u ∂x = ∂v ∂y ∂u ∂y =- ∂v ∂x . Since u and v already satisfy the Cauchy-Riemann equations, to show that v is conjugate to u it suffices to prove that v is also harmonic. Since u is harmonic, it’s second-order partial derivatives are all continuous (by definition). The Cauchy-Riemann equations above then imply the same is true of v ’s second-order partials. So all we need to do to prove that v is harmonic is verify that ∇ 2 v = 0. Appealing to the Cauchy-Riemann equations again we 1 have ∇ 2 v = ∂ 2 v ∂x 2 + ∂v ∂y 2 = ∂ ∂x ∂v ∂x + ∂ ∂y ∂v ∂y = ∂ ∂x- ∂u ∂y + ∂ ∂y ∂u ∂x =- ∂ 2 u ∂x∂y +- ∂ 2 u ∂y∂x = since the continuity of the second-order mixed partials implies their equality. Hence, v is harmonic. As noted above, this completes the proof. Notice that the only facts we used were that u and v were related by the Cauchy-Riemann equations and that u had continuous second-order partial derivatives. In fact, if these are the only assumptions that we make then we can, in fact, prove that both u and v are harmonic and that they are conjugate. Now on to the problem at hand. Since we are given that u is harmonic on the disk the claim shows that it is enough to verify that u and v satisfy the Cauchy-Riemann equations. To (hopefully) clarify things slightly, let me alter the notation of the problem. Let’s define: v ( x, y ) = c + Z y y ∂u ∂x ( x, t ) dt- Z x x ∂u ∂y ( s, y ) ds By the Fundamental Theorem of Calculus (the real-variable version) we have ∂v ∂y ( x, y ) = ∂ ∂y Z y y ∂u ∂x ( x, t ) dt- ∂ ∂y Z x x ∂u ∂y ( s, y ) ds = ∂u ∂x ( x, y ) since the second expression is not a function of y . This is one of the Cauchy-Riemann equations. Similar reasoning gives ∂v ∂x ( x, y ) = ∂ ∂x Z y y ∂u ∂x ( x, t ) dt- ∂u ∂y ( x, y ) ....
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- Fall '11
- dt, Logarithm, dz, cauchy-riemann equations