# Proof that, if $f$ and $|f|$ are analytic, then $f$ is constant

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.

• Analytic where? Connectivity has to enter into this somehow. – zhw. Dec 2 '18 at 19:12

$$|f|=|u+iv|=\sqrt{u^2+v^2}$$ 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, $$u^2+v^2\neq0$$

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

• I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0 – user601175 Dec 2 '18 at 17:19
• I edited the answer for the complete solution. – Shubham Johri Dec 2 '18 at 17:59
• could you explain the trivial case more rigorously please. what if the function is 0 in a non empty region – Dis-integrating Nov 30 at 21:28

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.

• How do you know that g is constant. Are all analytic functions with modulus <1 constant? – user601175 Dec 2 '18 at 17:08
• Any bounded entire function is constant (Liouville's Theorem) – Tito Eliatron Dec 2 '18 at 17:10

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.