Answer to Question #263786 in Complex Analysis for saduni

Question

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

Answer 1

Cauchy Riemann Equations:

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

$\frac{∂u} {∂x}= \frac{∂v}{ ∂y} \\and\\\frac{ ∂u}{ ∂y} =\frac{−∂v}{ ∂x}$

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

$Given\\f(z)=x^2+y^2\\U(x,y)=x^2+y^2=0\\U_x=2x;U_y=2y\\v_x=0;v_y=0$

Cauchy Riemann equations.

$U_x=v_y;U_y=-v_x\\2x=0;2y=0$

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

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

Is not analytic.