Prove that if f is an entire function which satisfies fz gre

Prove that if f is an entire function which satisfies |f(z)| greaterthanorequalto 1 on the entire plane, then f is constant.

Solution

Let f(z) = u (x,y) +i v (x,y).

Then |f(z)| = u² + v² =constant = c, where c is real number.                          .....(1)

Differentiating partially (1) with respect to x and y,

2 u u_x +2 v v _x = 0                                                                                                              ......(2)

2 u u_y +2 v v _y = 0                                        ......(3)

Since te function f(z) is entire function , it satisfies Cauchy-Riemann equations. Using the Cauchy-Riemann equations, i.e. u_x = v _y and    u_y = - v _x , and substituting in equations (2) and (3). Here we obtain the equations in terms of u_x and u_y :

2 u u_x - 2 v u _y = 0                                                                                                ......(4)

2 u u_y + 2 v u _x = 0                                                                                                                                                               ......(5)

Multiplying the (4) by u and (5) by v and adding,

(2 u u_x - 2 v u _y) u +(2 u u_y +2 v u _x) v = 0,

2(u² + v² ) u_x = 2 c u_x = 0,

It implies that u_x = 0, since c # 0. Hence, u( x, y) is independent of x.

Multiplying the (4) by v and (5) by u and subtracting,

(2 u u_x- 2 v u _y) v - (2 u u_y +2 v u _x) u = 0,

2(u² + v² ) u_y = 2 c u_y = 0,

It implies that u_y = 0, since c # 0. Hence, u( x, y) is independent of y.

Therefore, u(x,y) = real constant = c₁      ......(6)

Similarly, using the Cauchy-Riemann equations, i.e. u_x = v _y and    u_y = - v _x , and writing the equations in terms of v_x and v_y :

2 u v_y + 2 v v _x = 0                                                                                                                                                                ......(7)

- 2 u v_x + 2 v v _y = 0                                                                                                                                                             ......(8)

Multiplying the (7) by u and (8) by v and adding,

(2 u v_y + 2 v v _x ) u +(- 2 u v_x + 2 v v _y ) v = 0,

2(u² + v² ) v_y = 2 c v_y = 0,

It implies that v_y = 0, since c # 0. Hence, v( x, y) is independent of y.

Multiplying the (7) by v and (8) by u and subtracting,

(2 u v_y + 2 v v _x ) v - (- 2 u v_x + 2 v v _y ) u = 0,

2(u² + v² ) v_x = 2 c v_x = 0,

It implies that v_x = 0, since c # 0. Hence, v( x, y) is independent of x.

Hence, v(x,y) = real constant = c₂         ......(9)

Thus, f(z) = u (x,y) +i v (x,y) = c₁ + i c₂ = A complex constant.