Prove Constant: Entire Function f(z)=f(1/z)

[Amer]

if f is entire and f(z) = f(1/z) for all z, Prove f is constant
any hint ?

I was thinking about if I can prove that f' = 0 then we are done so
$f(z) = f\left(\dfrac{1}{z}\right)$
$f(x,y) = f\left(\dfrac{x}{x^2+y^2} , \dfrac{-y}{x^2+y^2} \right) $

I thought about Cauchy remain Equations.

Thanks

 

[Petek]

I think I have a solution that uses Taylor and Laurent series. If you know about those series, that's a hint. Otherwise, it might be helpful for you to state in which section of text this problem occurs.

 

[Opalg]

if f is entire and f(z) = f(1/z) for all z, Prove f is constant
any hint ?

Use Liouville's theorem – show that $f$ is bounded in the unit ball. Then the condition $ f(z) = f(1/z)$ shows that it is also bounded outside the unit ball, hence everywhere.

 

[Petek]

Here's my solution, which is quite different from Opalg's elegant argument:

Since $f$ is entire, we can write

$f(z) = \sum_{n=1}^{\infty}a_nz^n$

which converges for all $z \in \mathbb{C}$. Rewrite this as a Laurent series:

$f(z) = \sum_{n=-\infty}^{\infty}a_nz^n$

where $a_n = 0$ for $n<0$. Then

$f(1/z) = \sum_{n=-\infty}^{\infty}a_{-n}z^n$

Since $f(z)=f(1/z)$, we have

$\sum_{n=-\infty}^{\infty}a_nz^n=\sum_{n=-\infty}^{\infty}a_{-n}z^n$

By the uniqueness property of Laurent series, we see that $a_n=a_{-n}$ for all $n$. Therefore, $a_n=0$ for all $n \neq 0$. We conclude that $f(z)=a_0$ for all $z \in \mathbb{C}$, so $f$ is constant.

 

[Amer]

Here's my solution, which is quite different from Opalg's elegant argument:

Since $f$ is entire, we can write

$f(z) = \sum_{n=1}^{\infty}a_nz^n$

which converges for all $z \in \mathbb{C}$. Rewrite this as a Laurent series:

$f(z) = \sum_{n=-\infty}^{\infty}a_nz^n$

where $a_n = 0$ for $n<0$. Then

$f(1/z) = \sum_{n=-\infty}^{\infty}a_{-n}z^n$

Since $f(z)=f(1/z)$, we have

$\sum_{n=-\infty}^{\infty}a_nz^n=\sum_{n=-\infty}^{\infty}a_{-n}z^n$

By the uniqueness property of Laurent series, we see that $a_n=a_{-n}$ for all $n$. Therefore, $a_n=0$ for all $n \neq 0$. We conclude that $f(z)=a_0$ for all $z \in \mathbb{C}$, so $f$ is constant.

$\displaystyle f \left( \frac{1}{z} \right) = \sum_{n=-\infty}^{\infty} a_{-n} z^n $

Why if $a_n = a_{-n} $ we conclude $a_n = 0 $ for all $n$ non zero ?
Thanks

 

[Petek]

$\displaystyle f \left( \frac{1}{z} \right) = \sum_{n=-\infty}^{\infty} a_{-n} z^n $

Why if $a_n = a_{-n} $ we conclude $a_n = 0 $ for all $n$ non zero ?
Thanks

Sorry, I should have explained that better. In the expression $f(z)=\sum_{n=-\infty}^\infty a_nz^n$ all of the $a_n$ with $n<0$ are zero. On the other hand, in the expression $f(1/z)=\sum_{n=-\infty}^\infty a_{-n}z^n$ all of the $a_n$ with $n>0$ are zero. The equation $a_n=a_{-n}$, simply says that the coefficients of each power of $z$ in the two series are equal. Since, for nonzero $n$, the coefficient of $z^n$ is $0$ in at least one of the series, we conclude that all the coefficients of such powers of $z$ are $0$.

Hope that makes my argument more clear.

 

[Amer]

Sorry, I should have explained that better. In the expression $f(z)=\sum_{n=-\infty}^\infty a_nz^n$ all of the $a_n$ with $n<0$ are zero. On the other hand, in the expression $f(1/z)=\sum_{n=-\infty}^\infty a_{-n}z^n$ all of the $a_n$ with $n>0$ are zero. The equation $a_n=a_{-n}$, simply says that the coefficients of each power of $z$ in the two series are equal. Since, for nonzero $n$, the coefficient of $z^n$ is $0$ in at least one of the series, we conclude that all the coefficients of such powers of $z$ are $0$.

Hope that makes my argument more clear.

Very clear thanks for your effort.