Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.
Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.
I am not answering the question and inviting all users to answer it. It is a nice problem.
I will answer it within a week, but I am letting everyone try.
First of all it is trivial to check that the equation is satisfied trivially for $z=0$ since from the functional equation we have $$f(0)f(0)=0\implies f(0)=0$$
Now use $z\mapsto iz$ in the functional equation to get $$f(iz)f(-z)=-z^2$$ Adding this to the original equation we get $$\displaystyle f(iz)(f(z)+f(-z))=0$$
So either $f(iz)=0$ (which trivially gives that $f(z)=0$, $\forall$ $ z\in $ $\mathbb{C}$ and hence the result will be proved) or $\displaystyle f(z)+f(-z)=0$.
So in either case the last equality holds $\forall$ $z\in\Bbb{C}$
For $z=0$, we get $f(0)f(0)=0$ and so the claim holds for $z=0$. On the other hand, for $z\ne 0$, we necessarily have $f(z)\ne0$ (and $f(iz)\ne0$). From $$f(z)f(iz)=z^2$$ we get $$f(iz)f(-z) = f(iz)f(i^2z)=(iz)^2=-z^2,$$ so that $$ (f(z)+f(-z))f(iz) = z^2-z^2=0.$$ As $f(iz)\ne0$, the claim follows.
