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How to show that $f-g$ is imaginary constant in $\mathbb{D}$? Let $f$ and $g$ be continuous functions in $\bar{\mathbb{D}}$ and analytic in $\mathbb{D}$. Show that if $\mathfrak{R}f=\mathfrak{R}g$ at $\partial \mathbb{D}$, then $f-g$ is imaginary constant at $\bar{\mathbb{D}}$.

some hint, please.

I have found that if $f(z)=z^2$ and $g(z)=z^2-i$, then $\mathfrak{R}f=x^2-y^2=\mathfrak{R}g$ and $f-g=i$, which is imaginary constant $\forall z \in \bar{\mathbb{D}} \subset C$.

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Hint: $\Re f$, $\Re g$ are harmonic. Use Cauchy-Riemann. –  Ayman Hourieh Mar 11 '13 at 12:32
    
Aynab Hourieh: do you mean that use C-R equations on $u$ and $v$ of function $f=u+iv$? –  laovultai Mar 11 '13 at 12:41

2 Answers 2

up vote 2 down vote accepted

Hint: The real part of a holomorphic function is harmonic. So basically, if $u = \Re(f-g)$, a first step is to see why $u$ is zero on the whole of $\mathbb{D}$, $u$ being zero on the boundary of $\bar{\mathbb{D}}$.

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Hint: Maximum Modulus Theorem

More hint: look at the function $e^{f -g}$

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