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Suppose $f$ is analytic in the open unit disc and real valued on the radii $[0, 1)$ and $[0, e^{i \pi \sqrt 2})$. Prove that $f$ is constant. I'm not sure how to solve this.

alexwlchan
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user201756
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    Hint: Think about the Cauchy–Riemann equations. – SomeStrangeUser Dec 17 '14 at 07:45
  • I know that from the Cauchy-Riemann equations comes a theorem that says if f is analytic and real valued on a domain d, then f is constant. Since [0,1) and [0,$e^(ipisqrt(2))$) is contained within the unit disc and thus f meets the criteria of the theorem over this region and is constant. Is it really that simple, or am I misreading something? – user201756 Dec 17 '14 at 08:02
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    @user201756 That's not a domain - a domain is open. You need to think about taking the partial derivatives at $0$. –  Dec 17 '14 at 08:22
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    Hint: If $f$ is not constant, it's approximately $a_{0} + a_{k} z^{k}$ for small $z$ and some positive integer $k$. – gtrrebel Dec 17 '14 at 08:37
  • @user201756 Don't you satisfied with my answer? Should I make it more clear or finally you can accept it? – Jihad Dec 18 '14 at 18:44

2 Answers2

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Proof. Since $f$ is analytic we can write that $$ f(z) = \sum_{k=0}^\infty a_kz^k = a_0 + \sum_{k=1}^\infty a_kz^k $$ $a_0 = f(0) - $ real. Assume that some $\exists k_0>0:a_{k_0} \neq 0$. Then $$ f(z) = a_0 + a_kz^{k_0} + O(z^{k_0+1}) $$ Since it is real-valued on $[0,1)$ then $$ \lim_{x\rightarrow0}\frac{f(x) - a_0}{x^{k_0}} = a_{k_0} \in \mathbb{R}. $$ Since it is real-valued on $[0,e^{i\pi\sqrt{2}})$ then for $t \in [0, 1)$ $$ f(te^{i\pi\sqrt{2}}) = a_0+a_{k_0}t^{k_0}e^{i\pi\sqrt{2}k_0}+O(t^{k_{0}+1}) $$ For some $t_0$ this $|O(t_0^{k_0+1})| < Ct^{k_0}$ (I don't want to calculate this constant $C$, but it have to be less than $|a_{k_0}e^{i\pi\sqrt{2}}-a_0|$), hence $f(t_0e^{i\pi\sqrt{2}}) \notin \mathbb{R}$ and with have a contradiction with the assumption that $a_{k_0} \neq 0$.

Edit. $e^{i\pi\sqrt{2}n} \in \mathbb{C} \backslash\mathbb{R}$ since $\sqrt{2}$ is irrational.

Jihad
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  • You might want to mention why it is important for this question that $\sqrt2$ is irrational; why the result is not true if $\sqrt2$ is replaced by a rational number. – robjohn Dec 18 '14 at 00:32
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$f$ is analytic so we can write: $$ f(z) = \sum_{n=0}^\infty a_nz^n $$

Specifically for $x \in [0,1)$: $$ f(x) = \sum_{n=0}^\infty a_nx^n = \sum_{n=0}^\infty \text{Re}a_nx^n + i\sum_{n=0}^\infty \text{Im}a_nx^n $$

Since $f$ is real valued in $[0, 1)$ we get for $x \in [0,1)$: $$ \sum_{n=0}^\infty \text{Im}a_nx^n = 0 $$

This is a Tayolr series with real coefficients of the real $0$ function and so $\text{Im}a_n$ must all be $0$ (because of the uniqueness of the Taylor series) meaning that $a_n$ are all real numbers.

So we got that $f(z) = \sum_{n=0}^\infty a_nz^n$ and $a_n$ are real.

Now for $z \in [0,e^{i\pi\sqrt{2}})$ we write $z = te^{i\pi\sqrt{2}}$ where $t \in [0,1)$

And then: $$ f(z) = \sum a_nz^n = \sum a_n(te^{i\pi\sqrt{2}})^n = \sum a_nt^ne^{i\pi n\sqrt{2}} = $$ $$ \sum a_nt^n\cos\pi n\sqrt{2} + i\sum a_nt^n\sin\pi n\sqrt{2} $$

Since $f$ is real valued in $[0,e^{i\pi\sqrt{2}})$, and $a_n$ are real we get that for $t \in [0,1)$ $$ \sum a_n\sin\pi n\sqrt{2}t^n = 0 $$

And from here we get that $a_n\sin\pi n\sqrt{2} = 0$ for all $n$. For $n > 0$ it must be that $a_n=0$ but $a_0$ may or may not be $0$.

So in total we got that $f(z) = a_0$

benji
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