Question #263786

Show that 𝑓(𝑧) = 𝑥 2 + 𝑦 2 where 𝑧 = 𝑥 + 𝑖𝑦 is not analytic anywhere using Cauchy Riemann equations.


1
Expert's answer
2021-11-10T18:07:18-0500

Cauchy Riemann Equations:

Given, f(x+iy)=u(x,y)+iv(x,y)f(x+iy)=u(x,y)+iv(x,y) and it should satisfy 

ux=vyanduy=vx\frac{∂u} {∂x}= \frac{∂v}{ ∂y} \\and\\\frac{ ∂u}{ ∂y} =\frac{−∂v}{ ∂x}


If Cauchy Riemann equations are satisfied, then the function is analytic.


Givenf(z)=x2+y2U(x,y)=x2+y2=0Ux=2x;Uy=2yvx=0;vy=0Given\\f(z)=x^2+y^2\\U(x,y)=x^2+y^2=0\\U_x=2x;U_y=2y\\v_x=0;v_y=0

Cauchy Riemann equations.

Ux=vy;Uy=vx2x=0;2y=0U_x=v_y;U_y=-v_x\\2x=0;2y=0

Since the given function doesn't satisfy the property of Cauchy Riemann equation which shows

that f(z)=x2+y2f(z)=x^2+y^2\\

Is not analytic.


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