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# 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

### 2 Answers

- 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.

But

∂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.