Answer to Question #268331 in Complex Analysis for Ashis

Question

If f(z) is an entire function such that f(z)=f(-z) then there exists an entire function g(z) such that f(z)=g(z^2)

Answer 1

If a complex function is analytic at all finite points of the complex plane , then it is said to be entire.

let

$z=x+iy$

$f(z)=u(x,y)+iv(x,y)$

since $f(z)=f(-z)$ :

$u(x,y)=u(-x,-y),v(x,y)=v(-x,-y)$

then:

since $z^2=(x+iy)^2=(x^2-y^2)+ixy$ , then:

$g(z^2)=u_1(x^2-y^2,xy)+iv_1(x^2-y^2,xy)$

since $f(z)=g(z^2)$ , then:

$u_1(x^2-y^2,xy)=u(x,y)=u(-x,-y)$

$v_1(x^2-y^2,xy)=v(x,y)=v(-x,-y)$