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I am having trouble proving the following two questions about constant functions.

1) If $w=f(z)$ is an analytic function that maps all $z$ in region $D$ to a portion of a line, then $f(z)$ is constant.

2) If $f(z)=u(z)+iv(z)$ is an entire function such that $u(z)v(z)=3$ for all points in the complex plane, then $f$ is constant.

I understand that it is enough to prove that $f(z)$ is locally constant since the domain is a region. I tried using the fact that if the partial derivatives are both $0$, then $f(z)$ is locally constant and hence constant.

Thanks.

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  • $\begingroup$ For (1), have you met the open mapping theorem? $\endgroup$
    – Old John
    Dec 18 '13 at 21:57
  • $\begingroup$ No, I have not. What I think I should be using is the equivalence of the following statements. (i) f locally constant (ii) u and v locally constant (i.e.: 0 partial derivatives) (iii) f analytic and f'(z)=o for all z $\endgroup$
    – Maria
    Dec 18 '13 at 21:57
  • $\begingroup$ By a portion of a line do you mean a portion of the x-axis? $\endgroup$
    – user99680
    Dec 18 '13 at 21:59
  • $\begingroup$ OK, so we need a proof from elementary principles - I will have a think. $\endgroup$
    – Old John
    Dec 18 '13 at 21:59
  • $\begingroup$ @user99680 No, the question is in that general form. Had it been the x-axis only I would have used part(iii) in my previous comment and the problem would have been solved. $\endgroup$
    – Maria
    Dec 18 '13 at 22:03
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I think this may be in the right direction:

Assume $u(z)v(z)=3$ . Substituting back , we get $f(z)=u(z)+ i(3/v(z))$ (Notice we cannot have $v(z)=0$ , since $u.v=3$).

Now we apply Cauchy-Riemann:

i)$u_x= v_y= \frac {-3u_y}{u^2} $

ii)$u_y= -v_x = \frac {3u_x}{u^2}$

Now substitute $u_y$ in ii) into i) , to get: $u_x= \frac {-3(3u_x/u^2)}{u^2}=\frac{9u_x}{u^4}$ , and we get, $u_x(1-\frac {9}{u^4})=0$.

Similar argument for $v_x$.

For the second part, we multiply $f(z)$ by $e^{i\theta}$ , so that the line is rotated to the Real axis, and we set $g(z)=e^{i \theta}f(z)$, and let $g(z)=u+i.0$. Since $g(z)$ is analytic, we have:

$u_x= 0=v_y; u_y=0=v_x$, so that $u$ is constant . Then we translate back to the original function by the constant $e^{i (-\theta)}$, to get a constant function.

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For 1), as already suggested, use the Open Mapping Theorem: If $f$ is not constant, then it is open. In particular, $f(D)$ is open, and no part of a line is open in $\mathbb C$.

If you wish to avoid the use of the Open Mapping Theorem, the fact that the image of $f=u+iv$ is part of a line is equivalent to: There exist $a,b,c\in\mathbb R$, $(a,b)\ne (0,0)$, such that $au(z)+bv(z)=c$, for all $z\in D$. Thus $$ \mathrm{Im}\big(f(z)(b+ia)\big)=\mathrm{Im}\big( (u(z)+iv(z))(b+ia)\big)=au(z)+bv(z)=c, $$ and thus $f(z)(b+ia)$ is constant, whihc in turn implies that $f$ is constant.

For 2), then $\mathrm{Im}(f^2)=2uv=6$, which implies that $f^2$ is constant, and consequently, $f$ is also constant.

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  • $\begingroup$ Thanks for your answer. For (2), isn't $f^2=u^2+v^2+2iuv=u^2+v^2+6i$? $\endgroup$
    – Maria
    Dec 18 '13 at 22:24
  • $\begingroup$ Also I am expected to prove (1) without using Open Mapping Theorem $\endgroup$
    – Maria
    Dec 18 '13 at 22:26
  • $\begingroup$ I made the correction. $\endgroup$ Dec 18 '13 at 22:34
  • $\begingroup$ Thanks once more! Now I will try to prove that if $f^2$ is constant, $f$ is also constant. $\endgroup$
    – Maria
    Dec 18 '13 at 22:37
  • $\begingroup$ See also edited version of the answer. $\endgroup$ Dec 18 '13 at 22:41

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