Answer to Question #217163 in Calculus for vindhya

Question #217163

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
2021-10-29T03:00:37-0400

"\\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"



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