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Let $f(z) $ be an entire function such that for some constant $\alpha$, $|f(z)| \leqslant \alpha |z|^3 $ for all $|z| \geqslant 1$ and $f(z)=f(iz)$ for all $z$.

then comment about existence of such function.
My work $f$ will be polynomial of degree at most $3$ so $f(z) = a + bz + cz^2 + dz^3 \tag{1}$

$f(iz) = a + b(iz) + c(iz)^2 + d(iz)^3$
$f(z) = a +ibz - cz^2 -idz^3 \tag{2}$ comparing equation $1$ and $2$ we get $b=c=d=0$
hence $f(z)$ is constant polynomial.

Is this correct?

Paul Sinclair
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Saurabh Rana
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    $\alpha$ does not appear in the condition. – Gary Mar 14 '22 at 10:53
  • @Gary yes it was mistake on my part, i forgot to add it. i have corrected it. – Saurabh Rana Mar 14 '22 at 10:59
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    Perhaps you should justify your claim that $f$ will be a polynomial of degree at most $3$. – Gary Mar 14 '22 at 11:18
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    @Gary That has appeared too many times on MSE, so I think OP need not spell it out. Here is one instance: https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial?rq=1 – Kavi Rama Murthy Mar 14 '22 at 11:41
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    @KaviRamaMurthy I know why this is true, but perhaps this is a homework problem and the OP is required to post full solution. – Gary Mar 14 '22 at 12:33

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