I need a little help with this proof?
Prove that f(x) = x^3 is an entire function
Prove that g(x) - abs(x)^2 is not analytic in any domain of the complex plane
Any help would be greatly appreciated
- EulerLv 68 years agoFavorite Answer
dƒ/dz = 3z² exists and is continuous everywhere so ƒ(z) = z³ is analytic everywhere
(it is entire).
Setting z = x + i y, we have g(z) = x² + y²,
So the decomposition of g(z) into real and imaginary parts is
u(x + i y) = x² + y² and
v(x + i y) = 0.
If g(z) is differentiable at z = x + i y, then by the Cauchy Riemann equations,
∂u(x + i y)/∂x = ∂v(x + i y)/∂y and
∂u(x + i y)/∂y = -∂v(x + i y)/∂x.
∂u(x + i y)/∂x = 2x, ∂u(x + i y)/∂y = 2y,
∂v(x + i y)/∂x = 0, ∂v(x + i y)/∂y = 0.
So 2x = 0 and 2y = 0.
So the only point x + i y where g(x + i y) is differentiable is
x + i y = 0. So there is no open set on which g(z) is differentiable.
- M.j LimLv 78 years ago
Sorry, I can't proof that equation.