Answer to Question #218199 in Complex Analysis for Ahmad

Question #218199

if f(z) is an analytical function of z, then prove that ( d^2/dx^2 + d^2/ dy^2) log |f(z)|= 0


1
Expert's answer
2022-01-31T16:32:17-0500

We know that logf=12log(ffˉ)\log |f| = \frac{1}{2} \log(f\bar{f}). If we write f=u+ivf=u+iv, where u,vu, v are real functions, we find log(ffˉ)=log(u2+v2)\log(f\bar{f})=\log (u^2+v^2).

Therefore, we have

(2x2+2y2)logf=12(2x2+2y2)log(u2+v2)(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) \log|f| = \frac{1}{2} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\log(u^2+v^2) and let us calculate the derivatives :

122x2log(u2+v2)=x(uxu+vxvu2+v2)\frac{1}{2}\frac{\partial^2}{\partial x^2} \log(u^2+v^2)=\partial_x(\frac{u\partial_x u + v\partial_x v}{u^2+v^2}) taking the second derivative gives us a long expression :

(u2+v2)(uxxu+vxxv+(xu)2+(xv)2)2u2(xu)22v2(xv)24uvxuxv(u^2+v^2)(u\partial_{xx}u +v\partial_{xx} v+(\partial_xu)^2+(\partial_xv)^2)-\\ -2u^2(\partial_xu)^2-2v^2(\partial_xv)^2-4uv\partial_xu\partial_xv all divided by (u2+v2)2(u^2+v^2)^2

Developping this expression gives us

(u2+v2)(uxxu+vxxv)+(vxuuxv)2(uxu+vxv)2(u2+v2)2\frac{(u^2+v^2)(u\partial_{xx}u+v\partial_{xx} v) + (v\partial_x u - u\partial_x v)^2-(u\partial_x u + v\partial_x v)^2}{(u^2+v^2)^2}

Calculating the second derivative with respect to yy gives us a similar expression :

(u2+v2)(uyyu+vyyv)+(vyuuyv)2(uyu+vyv)2(u2+v2)2\frac{(u^2+v^2)(u\partial_{yy}u+v\partial_{yy} v) + (v\partial_y u - u\partial_y v)^2-(u\partial_y u + v\partial_y v)^2}{(u^2+v^2)^2}

Now let us use the Cauchy-Riemann equations (as ff is analytic, u,vu, v should satisfy them) :

{xu=yvyu=xv\begin{cases} \partial_x u = \partial_y v \\ \partial_y u = -\partial_x v \end{cases} and therefore (2x2+2y2)u=(2x2+2y2)v=0(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})u = (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})v=0

In addition, we have (uyu+vyv)2=(vxuuxv)2(u\partial_y u + v\partial_y v)^2 = (v\partial_x u - u\partial_x v)^2 and (uxu+vxv)2=(uyvvyu)2(u\partial_x u+v\partial_x v)^2 = (u\partial_y v - v\partial_y u)^2 by Cauchy-Riemann equations. Therefore, the sum

(2x2+2y2)logf=12(2x2+2y2)log(u2+v2)=0(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) \log|f| = \frac{1}{2} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\log(u^2+v^2)=0


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment