# Question involving the complex conjugate of a function.

The last problem on this released exam states:

Let $f$ be an entire function, and suppose that for all points $x$ on the real axis $f(x)$ is purely imaginary. Show that:

$$\overline{f(z)} = -f(\overline{z})$$

I tried using the hint given:

Let $g(z) = if(z)$ then $g(x)$ is real for points $x$ on the real axis. Then $\overline{g(x)} = g(x)$ since $g(x)$ is real. So,

$$\overline{if(x)} = if(x) \implies -i\overline{f(x)} = if(x)$$

And:

$$-i\overline{f(x)} = if(x) \implies \overline{f(x)} = -f(x) \implies \overline{f(x)} = -f(\overline{x})$$

So I know it holds for $x \in \mathbb{R}$ after that I'm not really sure what to do.

• – Seewoo Lee Sep 11 '16 at 1:52

If an entire function $g : \Bbb C \to \Bbb C$ is real on the real axis, then it satisfies $g(\overline{x}) = \overline{g(x)}$ for all $x \in \Bbb R$, as you noted. We claim it satisfies that equation in the entire complex plane. Your desired conclusion is obvious from there.

Proof of claim: Consider $\overline{g(\overline{z})}$. This function is also entire, and it agrees with $g$ on the real line. Therefore, we claim it must be equal to $g$ everywhere.

This is because if two holomorphic functions agree on a convergent sequence, they agree on their domains.

Alternatively, apply Claim #2 to $g(z) - \overline{g(\overline{z})}$.

Proof of Claim #2:

If an entire function is zero on the real line, then it is zero everywhere.

Let $h$ be an entire function that is zero on the real line.

Since $h$ is entire, it has a power series that is convergent everywhere. We'll center it at $0$ because it's convenient: $$h(z) = \sum_{n = 0}^\infty a_n z^n$$

There is a connection between derivatives and power series coefficients: $$a_n = \frac{h^{(n)}(0)}{n!}$$

Consider the derivatives of $h$. Recall that $$h'(z_0) = \frac{\partial u}{\partial x}(z_0) + i \frac{\partial v}{\partial x}(z_0)$$

Since both those terms are zero ($u$ and $v$ are constant $0$ on the real line), we have that $h'(x)$ is also zero for all $x \in \Bbb R$. Repeating this process, we get that $h^{(n)}$ is zero on the real line for all $n \in \Bbb N$. Therefore, $a_n = 0$.

Recall that $h$ agrees with its power series everywhere. Therefore, it is zero everywhere.

• Ah cool. My class hasn't covered the identity theorem. Guess the class that had this exam was a little ahead of us. Thanks. – Dair Sep 11 '16 at 1:55
• There's probably a more elementary approach, tbh, but I can't seem to find it at the moment :\ – Henry Swanson Sep 11 '16 at 1:56