Answer to Question #217163 in Calculus for vindhya

Question

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

Answer 1

$\implies f(z)=u(x,y)+iv(x,y)$

$logz=lnr+i\theta=ln\sqrt{(u(x,y))^2+(v(x,y))^2}+iarctan(v(x,y)/u(x,y))$

$\implies \frac{d}{dx}log|f(z)|=\frac{2uu_x+2vv_x}{2((u(x,y))^2+(v(x,y))^2)}+i\frac{1}{1+(v/u)^2}$

$\implies \frac{d}{dx}log|f(z)|=f_x/f$

$\implies \frac{d^2}{dx^2}log|f(z)|=\frac{f_{xx}f-f^2_x}{f^2}$

$\implies \frac{d}{dy}log|f(z)|=f_y/f$

$\implies \frac{d^2}{dy^2}log|f(z)|=\frac{f_{yy}f-f^2_y}{f^2}$

$\implies (\frac{d^2}{dx^2}+\frac{d^2}{dy^2})log|f(z)|=\frac{f_{xx}f-f^2_x}{f^2}+\frac{f_{yy}f-f^2_y}{f^2}$

$f(z)=u(x,y)+iv(x,y)$

for analytic function:

$u_x=v_y,u_y=-v_x$

$u_{xx}=-u_{yy},v_{xx}=-v_{yy}$

then:

$(\frac{d^2}{dx^2}+\frac{d^2}{dy^2})log|f(z)|=\frac{f\cdot(u_{xx}+iv_{xx}+u_{yy}+iv_{yy})-((u_x+iv_x)^2+(u_y+iv_y)^2))}{(u+iv)^2}$

$u_{xx}+iv_{xx}+u_{yy}+iv_{yy}=u_{xx}-iv_{yy}-u_{xx}+iv_{yy}=0$

$(u_x+iv_x)^2+(u_y+iv_y)^2=(u_x)^2+2iu_xv_x-(v_x)^2+(u_y)^2+2iu_yv_y-(v_y)^2=$

$=(v_y)^2-2iv_yu_y-(u_y)^2+(u_y)^2+2iv_yu_y-(v_y)^2=0$

So,

$(\frac{d^2}{dx^2}+\frac{d^2}{dy^2})log|f(z)|=0$