Find an example of an entire function such that $f(z)=-f(iz)$ for all $z$ Is there a systematic way to find an example of an entire function $f$ such that $f(z)=-f(iz)$ for all $z$? By testing at monomials, we find that $f(z)=z^6$ is a solution. But how can we find other solutions?
I tried writing $f=u+iv$ and then differentiated and then applied Cauchy-Riemann equations, but that just ended up with Cauchy-Riemann equations for $u$ and $v$ itself.
 A: Since $f$ is entire, we may expand it at $z = 0$: $$f(z) = \sum_{n \geq 0}a_nz^n,$$ where each $a_n$ is a complex number.
It follows that $$-f(iz) = \sum_{n\geq 0}-a_ni^nz^n$$ and the needed identity is thus equivalent to $$a_n = -a_ni^n, \forall n\geq 0.$$
This holds if and only if $a_n = 0$ for all $n$ such that $n\not\equiv 2\mod4$.
Therefore we may write $f(z) = a_2 z^2 + a_6 z^6 + a_{10}z^{10} + \cdots$, i.e. $f(z) = z^2g(z^4)$, where $g$ is an entire function.
A: *

*All the monomials $z^2,z^6,z^{10},\dots$ are solutions.

*If $g(z)$ is any entire function, then
$$
f(z) = \tfrac14\big( g(z) - g(iz)+g(-z)-g(-iz) \big) \tag{a}
$$
is also entire and satisfies $f(z)=-f(iz)$. (This is similar to how the combination $\frac12\big( h(x)-h(-x) \big)$ produces an odd function for any real-valued $h$.)

*If one applies the averaging operation (a) to a monomial $z^k$, then the result is $z^k$ again if $k\equiv 2\pmod 4$ and $0$ otherwise. In particular, the operation (a) converts the power series for $f(z)$ into the subseries consisting only of the terms $a_kz^k$ where $k\equiv 2\pmod 4$.

*Similarly, if $h(z)$ is any entire function, then $f(z)=z^2h(z^4)$ is an entire function with the given property, because its power series consists only of terms $a_kz^k$ where $k\equiv 2\pmod 4$.

