Thread: Absolute Values of Complex Functions

If f and g are continuous functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.

What is the analog for functions of a complex variable?

This is my suspicion: if f and g are nonvanishing analytic complex-valued functions on some connected subset D of C, and |f(z)| = |g(z)| for all z in D, what can be said of f and g? Is it necessary that f = cg, where |c| = 1? If this is right, how would one prove it?

Thanks in advance for any help.

Originally Posted by joeyjoejoe

If f and g are functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.

What is the analog for functions of a complex variable?

This is my suspicion: if f and g are nonvanishing complex-valued functions on some connected subset of C, and |f(z)| = |g(z)| for all z in D, what can be said of f and g? Is it necessary that f = cg, where |c| = 1? If this is right, how would one prove it?

Thanks in advance for any help.

$\displaystyle f(z)=e^{iRe(z)}\,,\,\,g(z)=1$ . It is always true that $\displaystyle |f(z)|=|g(z)| \,\,\forall z\in\mathbb{C}$ , but none is a scalar multiple of the other one, of course.

Tonio

Originally Posted by joeyjoejoe

If f and g are functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.

What if $\displaystyle g(x) = \begin{cases} f(x), & \mbox{if } x \geq 0 \\ -f(x), & \mbox{if } x < 0. \end{cases} $.

Here we have $\displaystyle |f(x)|=|g(x)| $ but $\displaystyle f(x)\neq\pm g(x) $.

I meant to say f and g are continuous (real variable case) and analytic (complex variable case) respectively.

I don't think that counterexample you give is analytic, so I'm still thinking it could be true.