holomorphic functions, roots of unity and harmonic numbers

If $$f$$ is a non-constant holomorphic function such that, for all $$z \in \mathbb{C}$$, exists a $$c \in \mathbb{C}$$ where $$f(cz) = f(z),$$ then $$c$$ must be a $$n$$-th root of unity, or there exists some counterexample?

Furthermore, there exists $$f$$ holomorphic function and $$n \in \mathbb{N}$$ such that $$f(z\mathcal{H}(n)) = f(z), \forall z \in \mathbb{C}?$$

In this case, $$\mathcal{H}(n)$$ denotes the $$n$$-th harmonic number: $$\mathcal{H}(n) = \sum\limits_{k = 1}^n \frac{1}{k}$$

• Do you mean the $c$ does not depend on $z$? – Berci Nov 15 '18 at 22:30
• @Berci Yes, $c$ does not depend on $z$. – 674123173797 - 4 Nov 15 '18 at 23:54

Regarding your first question: If you allow $$c$$ to depend on $$z$$, there is very little one can say about $$c$$.
Consider $$f(z) = \sin z$$. Given $$z \ne 0$$, set $$c = \frac{z + 2 \pi}{z}$$. For $$z = 0$$, set $$c = 2$$. Then $$f(cz) = f(z)$$ for all $$z$$.
Assume now that $$\exists c \, \forall z \, f(z) = f(cz) = f(c^{-1}z)$$ and $$f$$ is not constant and entire. Then $$c$$ must be a root of unity.
Clearly $$|c| = 1$$, since otherwise we have the sequence $$c^k$$ or $$c^{-k}$$ converging to $$0$$ on which $$f$$ is constant, implying that $$f$$ is constant. If $$c$$ is not a root of unity, then $$\{c^n : n \ge 0\}$$ is dense on the unit circle and therefore $$f(z) = f(1)$$ on the unit circle, again implying that $$f$$ is constant.
For the second part: replace $$z$$ by $$\frac z {H(n)}$$ to get $$f(z)=f(\frac z {H(n)})$$. By iteration $$f(z)=f(\frac z {H(n)^{k}})$$ for any $$k \geq 1$$. Letting $$k \to \infty$$ we get $$f(z)=f(0)$$ for all $$z$$ provided $$n >1$$. For $$n=1$$ there is no hypothesis, so $$f$$ can be any entire function.