Is $f(z) = e^{iz}$ is a bounded or unbounded function on $\Bbb C$?

Question

let $f:\Bbb C→\Bbb C$ defined by

$f(z) = e^{iz}$. where $\Bbb C$ is complex field

My question is whether $f(z) = e^{iz}$ is a bounded or unbounded function on C ?

My attempt : I take $f(z) = e^{iz}$

$f(x +iy) = \dfrac1{e^y}(\cos x +i\sin x)$ where $z = x+iy$

I know that sine and cosine is a bounded function,,,,,

Therfore $f(z) = e^{iz}$ is a bounded function on $\Bbb C$.

Is my answer is correct or not ? please verify it.

Answer 1

The function is unbounded because, for each $n\in\mathbb N$, $f(-ni)=e^n$.

Answer 2

$e^{iz}$ is unbounded on $\mathbb C$ as is every nonconstant entire function (by Liouville's Theorem).

In your statement, the factor $(\cos x + i\sin x)$ is indeed bounded (its absolute value is $1$), but the factor $1/e^{y}$ can be made arbitrarily large by taking $y$ to be a sufficiently large negative number.

Thus their product, which is $e^{iz}$, is unbounded.

Answer 3

Your function is indeed unbounded, as most comments have told you. However, something important to keep in mind is that you have just shown that $e^{iz}$ is a bounded function when restricted to the upper open complex plane $\mathbb{C}_+= \{z \in \mathbb{C}: \Im(z)>0 \}$. This can be useful sometimes.