# Find all entire functions $f$ such that $f^{(n)}(z) = z$ for all $z$, $n$ being a given positive integer

Find all entire functions $f$ such that $f^{(n)}(z) = z$ for all $z$, $n$ being a given positive integer.

I can not think such a function exist or not.can somebody help me please

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What is $f^n(z)$ ? (derivative or power or what ?) –  Jean-Claude Arbaut Mar 13 '13 at 17:29
What does $f_n$ mean? (@arbautjc jinx.) –  Alexander Gruber Mar 13 '13 at 17:29
Do you mean $f^n$ or $f^{(n)}$? To me, $f^n$ is the $n$-th iteration of $f$, e.g. $f^2 \equiv f \circ f$, while $f^{(n)}$ is the $n$-th derivative, e.g. $f^{(2)} \equiv f''$. –  Fly by Night Mar 13 '13 at 17:29
extremely sorry for my mistake.corrected it now. –  user59908 Mar 13 '13 at 17:31
I am also not sure as it mentioned in a question paper and nothing mentioned more than that. can we do this problem for both the cases that is for derivative and as power. –  user59908 Mar 13 '13 at 17:31

Since $f$ is entire, it has a series development $$f(z) = a_0 + a_1 z \ + ...$$ But then equation $f^{(n)}(z)=z$ implies $$f(z) = a_0 + a_1 z + ... + a_{n-1} z^{n-1} + {z^{n+1} \over {(n+1)!}}$$

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Hint: $$f^{(n+1)}(z)=\frac{d f^{(n)}}{dz} (z)=\frac{dz}{dz}=1$$

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Lemma. If $(g-h)'(z)=0$ then $(g-h)(z)=c$ so $g(z)=h(z)+c$.

Hint. Use the above lemma along with an inductive argument to show that $f$ must be a polynomial of degree $n+1$.

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Consider the example $f'''(z) = z$. Integrating gives $f''(z) = \frac{1}{2}z^2+c_1$ and in turn:

$$f'(z) = \frac{1}{6}z^3+c_1z+c_2 \, .$$

Integrating a final time gives:

$$f(z) = \frac{1}{4!}z^4+\frac{c_1}{2}z^2+c_2z+c_3 \, .$$

In general, if $f^{(n)}(z)=z$ then we can integrate $n$ times to give ourselves:

\begin{array}{ccc} f(z) &=& \frac{1}{(n+1)!}z^{n+1}+\frac{c_1}{(n-1)!}z^{n-1}+\frac{c_2}{(n-2)!}z^{n-2}+\cdots + c_{n-1}z+c_n \\ f(z) &=& \frac{1}{(n+1)!}z^{n+1} + b_1z^{n-1}+b_2z^{n-2}+\cdots+b_{n-1}z+b_n \end{array}

These are all polynomials, and so are entire. You can choose the constants $b_i \in \mathbb{C}$ however you like. Obviously, the simplest case would be $b_i = 0$ for all $i$ and hence:

$$f(z) = \frac{1}{(n+1)!}z^{n+1} \, .$$

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