Show that an analytic function with constant modulus is itself a constant
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See [Maximum Modulus Principle](http://en.wikipedia.org/wiki/Maximum_modulus_principle). It is impossible to know what you need in addition without you being more detailed in your question. – quid May 12 '15 at 14:41

Hello, welcome to Math Stack Exchange. Please be more detailed in your question and show your attempt. This will help us to give the best answers at your level and form that you will learn most. You can edit your post to add this. – wythagoras May 12 '15 at 14:45
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Hint: Write $f = u+iv$. If $f = 0$ then $f = 0$ and there's nothing to do. Otherwise, $u²+v² = c²$ for some constant $c > 0$. Differentiate implicitly w.r.t. $x$ and $y$ and use the CauchyRiemann equations to conclude that $u_x = u_y = v_x = v_y \equiv 0$.
Ivo Terek
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