Proving Analytic Function Convergence Using Constant Sum of Squares

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Homework Statement


Let f1(z),... fn(z) n >1 be analytic functions on a domain D such that |f1 (z)|2 +...+|fn(z)|2 = k (k is constant). Verify that all the functions must reduce to constants


Homework Equations





The Attempt at a Solution


Having the hardest time proving this, not sure where to being...
 
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What tools do you have to work with? I can probably think of a few ways to do this. Do you know what a harmonic function is?
 
You mean harmonic function in terms of being analytic, or something else?

Also i was given the hit that |fk(z)|2 (|f1(z)|2Delta f1(z)|2)
 
Well a harmonic function can be defined independently of analytic function theory, but yes the real and complex parts of an analytic function are harmonic, i.e., they satisfy http://en.wikipedia.org/wiki/Laplace's_equation#Laplace_equation_in_two_dimensions".

The original equation can be written in terms of the real and complex parts of each f_i (i = 1, ..., k). Differentiating the original equation twice with respect to x gives a new equation, call it (1). Differentiating the original equation twice with respect to y (use symmetry here), we get another equation (2). What happens when you add equations (1) and (2)?
 
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So would (1) be \partialx (|f1(z)|2 +...|fk(z)|2?
 
Well no, you should have two x's in the subscript, but never mind that. Writing it that way doesn't do anything since you didn't actually take derivatives. Put f_j = u_j + iv_j, (j = 1, ..., k) and write |f_1(z)|^2 +...|f_k(z)|^2 in terms of the real and complex parts of each f_j. Then take two partial derivatives with respect to x and call that (1).
 
Ok so i got |f1(z)|2 = f1(z)f1(z(bar)) = (x1+iy1)(x1-iy1) + ... + (xn+iyn)(xn-iyn)

then separated it into real and imaginare parts
Re{(x1)(x1) + ...+ (xn)(xn)
partial x = 2(x1)+ 2(x2) +...+ 2(xn)
or did you mean partial x = 2(x1)(dx/dz)+ 2(x2)(dx/dz) +...+ 2(xn)(dx/dz)? or do i need to keep it (x+iy)(x-iy) and then do the partial?
 
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