Regarding equality of the exponential of two complex numbers

Question

Let $z_1$ and $z_2$ be two complex numbers. I know that if $\exp(z_1)=\exp(z_2)$ may not imply $z_1=z_2$. But if $\exp(tz_1)=\exp(tz_2)$ for every $t\in \mathbb{R}$ does that imply $z_1=z_2$? If yes, then how?

Answer 1

Differentiate both sides with respect to $t$ to obtain

$$z_1\exp(tz_1)=z_2\exp(tz_2).$$

Setting $t=0$ completes the proof.

Answer 2

$\exp(z_1)=\exp(z_2)$ does not imply $z_1 = z_2$ but it implies $z_1 - z_2 = k2\pi i$ for some $k \in \mathbb{Z}$.

So, $\exp(tz_1)=\exp(tz_2)$ for every $t\in \mathbb{R}$ implies $z_1 - z_2 = \frac{k2\pi i}{t}$ for all $t \neq 0$. So, we can prove that the difference is smaller (closer to zero) than any non-zero number and hence it must be zero.

Answer 3

If $\exp(tz_1)=\exp(tz_2)$ for all t, then to each $t$ there is $k(t) \in \mathbb Z$ such that

$tz_1-tz_2= 2 k(t) \pi i$.

Hence

$(*)$ $|t||z_1-z_2|=2 |k(t)| \pi$.

Now assume that $z_1 \ne z_2$. $(*)$ shows that $k(t) \to 0$ for $t \to 0$. Since $k(t) \in \mathbb Z$, we get a $\delta >0$ such that $k(t)=0$ fot $0<|t|< \delta$. But then we have $z_1=z_2$ .....