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The question is this:

Let $\Omega$ be a domain in the upper half plane $\mathbb{H} = \{ z \in \mathbb{C} ~|~ \Im~ z > 0 \}$. Let $f,g \in O(\Omega)$ holomorphic, such that $f(i) = 0~~$, $f$ is real-valued, and

$$ |f(z)|^2 + 2|g(z)|^2 = |z|^2$$

for all $z \in \Omega$. Determine $f$ and $g$.

This is my attempt at a solution:

let $z = i$, then $f(z) = 0$, and $ 2|g(z)|^2 = |z|^2 = |i|^2 = 1$

Therefore, $$ |g(i)| = 1/2~~~\star$$

Then consider $z \in \Omega$ such that $z \neq i$, then:

$$|f(z)|^2 + 2|g(z)|^2 = |z|^2$$ $$f \bar{f} + 2 g \bar{g} = z \bar{z}$$ $$\frac{\partial }{\partial \bar{z}}(f \bar{f}) + 2\frac{\partial}{\partial \bar{z} }(g \bar{g}) = \frac{\partial}{\partial \bar{z}}(z \bar{z})$$ $$f\frac{\partial \bar{f}}{\partial \bar{z}} + \bar{f}\frac{\partial f}{\partial \bar{z}} + 2(g\frac{\partial \bar{g}}{\partial \bar{z}}+ \bar{g} \frac{\partial g}{\partial \bar{z}}) = z$$

The first 2 terms are $0$ since $f$ is real valued so $f = \bar{f}$ and since $f$ is holomorphic, the derivative with respect to $\bar{z}$ is also $0$. Also the last term is $0$ since $g$ is holomorphic.

So we are left with : $$ 2g\frac{\partial \bar{g}}{\partial \bar{z}} = z$$ $$2g \overline{\frac{\partial g}{\partial {z}}} = z$$

Then $$2gg'(z) = z~~~\star \star$$

This is where I am stuck. I don't know what to do with the information on $g$ and if I find it I can find f but I don't know how to proceed from here.

Any help would be appreciated.

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    $f$ is real-valued and holomorphic. What does that tell you? – 2017-02-11
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    that it's constant?... but how does that help? – 2017-02-11
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    Right. So with $f(i) = 0$ we have $f\equiv 0$ and the problem reduces to finding $g$ such that $2\lvert g(z)\rvert^2 = \lvert z\rvert^2$ for all $z \in \Omega$. What does that then tell you about $g(z)/z$? – 2017-02-11
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    its magnitude is 1/\sqrt2 – 2017-02-11
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    $1/\sqrt{2}$ actually, you forgot the squares. And a holomorphic function with constant modulus … – 2017-02-11
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    is also constant and equal to $1/\sqrt{2}$. so $|g| = 1/\sqrt 2 |z|. letting z = x + iy, $g = +/- 1/\sqrt{2} z$ – 2017-02-11
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    _Every_ $c$ with $\lvert c\rvert = 1/\sqrt{2}$ gives a possible $g$ per $g(z) = cz$. So far, no conditions have been given that would imply that $c$ is real. – 2017-02-11
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    so i just define a complex number c whose magnitude is $1/\sqrt 2$, then for each c, g = cz..am i right? – 2017-02-11
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    What does ${\partial \over \partial \bar{z}}$ mean? Is this the same as $\bar{\partial}$ in Rudin's Real & Complex Analysis? – 2017-02-11
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    I'm not sure about what's present in Rudin, but it's one of the Wirtinger derivatives. if $\frac{\partial f}{\partial \bar{z}} = 0$, then f is holomorphic – 2017-02-11
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    @DanielFischer thanks for the help – 2017-02-11
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    I guess it must be the Wirtinger derivative (new term to me). – 2017-02-11

1 Answers 1

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The question is more interesting without assuming $f$ real valued.

For $z\neq 0$ you have:

$$ \left| \frac{f(z)}{z} \right|^2 + 2 \left| \frac{g(z)}{z} \right|^2=1 $$

Pick a point $w$ in (the interior of) your domain and let $\theta,\phi$ be reals such that $e^{i\theta} f(w)/w\geq 0$ and $e^{i\phi} g(w)/w\geq 0$. Let $$F(z) = e^{2i\theta} \left(\frac{f(z)}{z}\right)^2 + 2e^{2i\phi} \left( \frac{g(z)}{z} \right)^2.$$ Then for any $z$, taking abs values: $|F(z)| \leq 1 = F(w)$ so the holomorphic function $F$ attains its maximum at the interior point $w$, whence must be constant (=1). Now, the equality $ |u|+|1-u|=1$ implies that $0\leq u\leq 1$ so the two functions in $F$ must be real, whence constant (since holomorphic). Therefore, $ f(z)=az$ and $g(z)=bz$ with $|a|^2+2|b|^2=1$. Finally, if $f$ has a root $a=0$.

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    +1: Nice observation: $|z|+|1-z| = 1$ $\implies $ $z \in [0,1]$. – 2017-02-13