Question

show that an analytic function with constant argument is constant​

Answer 1

Answer:

Write f(z) =u(x,y) + i.v(x,y). As f(z) is analytic, by Cauchy-Riemann conditions, u_x = v_y and u_y = -v_x, where the subscripts denote partial differentiation. By hypothesis, |f(z)|^2 = u(x,y)^2 + v(x,y)^2 = k (constant). Differentiating partially with respect to x and y , 2u.u_x+2v.v_x=0 and 2u.u_y+2v.v_y=0, which give the pair of linear equations: u.u_x -v.u_y=0 and v.u_x +u.u_y =0, (by use of the C.R. equations quoted above) in unknowns u_x and u_y and whose determinant of coefficients is ∆= u^2 +v^2 =k for every x and y. If k=0, then u(x,y) = 0 = v(x,y) which gives that f(z)=0 for every z and so f(z) is the 0 constant function. If k is nonzero, then the above homogeneous linear system of equations in u_x and u_y has only the trivial solution i.e. u_x and u_y are 0 everywhere, i.e u(x,y)=a, some constant. Similarly we can frame a pair of linear equations in v_x and v_y and conclude that v(x,y) =b(say). Hence f(z) = a + i.b is a constant function