# A function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$ satisfies $f(z)+f(-z)=0$.(MADHAVA-2020)

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.