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I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.

My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.

Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.

Thanks in advance

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    $\begingroup$ Analytic where? Connectivity has to enter into this somehow. $\endgroup$ – zhw. Dec 2 '18 at 19:12
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$|f|=|u+iv|=\sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.

$\implies |f|=U+iV; U=\sqrt{u^2+v^2}, V=0$

$U_x=V_y\implies uu_x+vv_x=0$

$U_y=-V_x\implies uu_y+vv_y=0$

From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$

$\implies uu_x-vu_y=0$

$\implies uu_y+vu_x=0$

$\implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$

For non-trivial solution, $u^2+v^2\neq0$

$\implies u_x=u_y=v_x=v_y=0$

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  • $\begingroup$ I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0 $\endgroup$ – user601175 Dec 2 '18 at 17:19
  • $\begingroup$ I edited the answer for the complete solution. $\endgroup$ – Shubham Johri Dec 2 '18 at 17:59
  • $\begingroup$ could you explain the trivial case more rigorously please. what if the function is 0 in a non empty region $\endgroup$ – Dis-integrating Nov 30 at 21:28
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If $f$ and $|f|$ are both analytic (in the whole complex plane $\mathbb C$), $g:=\frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=\frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=c\left(1+\frac{|c|}{1-|c|}\right)$ which is a constant.

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  • $\begingroup$ How do you know that g is constant. Are all analytic functions with modulus <1 constant? $\endgroup$ – user601175 Dec 2 '18 at 17:08
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    $\begingroup$ Any bounded entire function is constant (Liouville's Theorem) $\endgroup$ – Tito Eliatron Dec 2 '18 at 17:10
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Assume $f,|f|$ are analytic in the connected open set $U.$ If $f\equiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)\subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= f\bar f$ is analytic. Thus in $D(a,r),$ $(f\bar f)/f = \bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.

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