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I am working on the question:

Suppose $f$ is entire, with real and imaginary parts $u$ and $v$ satisfying $u(z) v(z) = 3$ for all $z$. Show that $f$ is constant.

I understand that I must show $f'(z) = 0$. The first thing I did here was to set up the Cauchy Riemann Equations: $u_x = v_y$, and $u_y = -v_x$. However, I am not sure how to use $u(z) v(z) = 3$. Clearly I have to take some sort of derivative, but I'm not very familiar with how partial derivatives work. Can I take the partial derivative of both sides of $u(z) v(z) = 3$ and say $u_x (z) v(z) + u(z) v_x(z) = 0$? If yes, how should I proceed afterwards? I have worked under this assumption, but I still didn't get the desired $f'(z) = 0$.

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    Consider $f^2$. – 2017-02-22
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    More generally, the [open mapping theorem](https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)) takes care of all exercises of this type. – 2017-02-22

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If $f(z)$ is entire then so is $f(z)^2$, and if $f=u+iv$ then $$ f^2=(u+iv)^2=u^2-v^2+2iuv$$

The condition $u(z)v(z)=3$ therefore implies that $f(z)^2$ has a constant imaginary part, so now it's enough to show that an entire function with a constant imaginary part is constant.

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    Thanks, very nice method! Do you think there is also a way to solve it witouth squaring by differentiating $uv = 3$? – 2017-02-22
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    I don't see immediately how to make that work, but I wouldn't rule out the possibility that it could be done. – 2017-02-22
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    Hm okay. Is it correct to apply the product rule to $uv = 3$ and say $u_x v + u v_x = 0$? I am not too familiar with the rules of partial derivatives. – 2017-02-22
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    Yes, that's fine. – 2017-02-22
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    Ok thank you very much! – 2017-02-22
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    You have to use the Cauchy Riemann equations to see the last part right (entire function with constant img is constant)? – 2017-02-22
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    Yes, that's correct. – 2017-02-23
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Note that if $f=u+iv$ and $v=3/u$, then the Cauchy-Riemann Equations reveal that

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\implies u_x=-3u_y/u^2 \tag 1$$

and

$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\implies u_y=3u_x/u^2 \tag 2$$

Using $(1)$ and $(2)$ together yields

$$u_x^2+u_y^2=0\implies u=\text{constant}$$

Hence $f$ is constant.

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    Thank you Ovi! Much appreciative. -Mark – 2017-02-22
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    Is there a clear way to see why $u_x^2 + u_y^2 = 0 \implies u$ is constant? – 2017-02-22
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    @hawk If the sum of two non-negative number is zero, what is the value of those numbers? – 2017-02-22
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    If they are real, then $0$. But can't have $u_x = \pm i u_y$? – 2017-02-22
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    @hawk By assumption, $u$ is real. – 2017-02-22
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    Oh right, we always assume real $u$ and $v$ when we write $u + iv$, otherwise that generates a lot of confusion. – 2017-02-23
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    @hawk Yes. You have it now. Well done. – 2017-02-23
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Consider the function $\displaystyle g(z)=e^{if^2(z)}$. Then , $g$ is an entire function.

Now , $\displaystyle |g(z)|=\left|e^{i(u^2-v^2)}\right|.\left|e^{-2uv}\right|=e^{-6}$ , for all $z\in \mathbb C$. So , $g$ is bounded in $\mathbb C$.

So by Liouville's theorem $g$ is constant and hence $f$ is constant.

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Differentiating partially w.r.t $x$ and $y$ separately, you get $uv_x+vu_x=0$ and $uv_y+vu_y=0$. Use C-R equations to get $-uu_y+vu_x=0$ and $uu_x+vu_y=0$. On solving for $u_x$ you get

$(u^2+v^2)u_x=0\implies u_x=0$ or $u=constant$ as $(u^2+v^2)\ne 0$ unless $u=v=0$