Prove that $f(z) $ is an odd function when $z \in \mathbb{R}$

Question

Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function i,e $f(-z)= -f(z)$

My attempt : Here two cases will be possible

$1.$ when $z= x$

$2.$ when $z = ix $

I have no any issue when $z=ix$ and solution is alreaday given here

but i have some issue when $z=x$

$$f(x) = \overline{ f(\bar x) }$$

$$\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\bar a_nz^n \implies a_n=\bar a_n$$

\begin{align} f(x)=\sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_{2n}x^{2n}+\sum_{n=1}^{\infty}a_{2n+1}x^{2n+1} \end{align}

Here $f(x)$ is both even and odd

so here we are getting contradiction that $f(z)$ is odd that mean $f(x)$ is not even function

Answer 1

On the solution here

The guy first assumed z is only imaginary ($z = iy$). Then he used that to show that all even $a$'s should be zero. That is $a_{2n}=0$. This means $$f(z) = a_1z^1 + a_3z^3 + a_5z^5 + ...$$ for all $z = x +iy $

Then he used that to show that the function is odd for any $z$ ($z = x+iy$)

That is: $$ f(-z) = a_1(-z)^1 + a_3(-z)^3 + a_5(-z)^5 + ... = - (a_1z^1 + a_3z^3 + a_5z^5 + ...) = -f(z) $$

Answer 2

From this answer we have show that

$a_{2n}=0$ for all $n$.

now $f(x+iy)= \sum_{n=0}^{\infty}a_n(x+iy)^n= a_0+a_1(x+iy)^1 + a_2(x+iy)^2+....... $ give odd function