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Recall that a function $f(z)$ is called radial if it is constant along the circles of center $0$. Let $f$ be a radial holomorphic function defined on the unit disc $D$.

Show that $f$ is constant. (Hint: apply the Cauchy-Riemann equations)

I've come across this problem but I've never heard of radial functions.

I know that if $f$ is holomorphic then it is complex differentiable and so $\frac{\delta f}{\delta x} $ and $\frac{\delta f}{\delta y}$ both exist. And so we have the Cauchy-Riemann equations that hold.

But how do I use these facts (that $f$ is radial and holomorphic) to prove that it's constant ?

Please I need help. Thank you

JOJO
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1 Answers1

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Use Cauchy Riemann equations in polar coordinates $(r,\theta)$ [ Ref: Cauchy-Riemann equations in polar form. ]

Since the real and imaginary parts depend only on $r$ and not on $\theta$ the partial derivatives of $u =\Re f$ and $v =\Im f$ w.r.t. $\theta$ are $0$. But Cauchy Riemann equations now show that the partial derivatives w.r.t. $r$ are also $0$ forcing these functions to be constants.

Kavi Rama Murthy
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  • What do you mean by "the real and imaginary parts depend only on r and not on θ the partial derivatives of u=Rf and v=If w.r.t. θ are 0". How do the real and imaginary parts depend on r only? – JOJO Feb 13 '20 at 12:18
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    The hypothesis is $f(z)=f(|z|)$ for all $z$. [ Note that $z$ and $|z|$ lie on the same circle centered at the origin]. Hence $f(re^{i\theta})=f(r)$ which does not depend on $\theta$. – Kavi Rama Murthy Feb 13 '20 at 12:24