Proof Verification: No $x$ such that $e^x$ = 0

Question

In my school's grade 10 curriculum we learnt something called the natural logarithm. And of course, comes $ln(0) = undefined$.

Now of course, every generic math teacher should ask their students, "Why?"

Here's my try: $$e^{a + bi} = e^a \times e^{bi} = 0$$and since $e^a > 0$, $$e^{bi} = 0$$ which implies that $$e^{(\pi\times\frac{b}{\pi}\times i)}=0$$ Let $c = \frac{b}{\pi}$, then $$(e^{\pi i})^{c} = \cos(\pi c) + i \sin (\pi c) = 0. $$

thus $$\cos(\pi c) = -i \sin (\pi c)$$ and since $\cos(x), \sin(x)$ for all real $x$ is real, $$\cos (\pi c) =-i \sin(\pi c)=0$$which has no solutions.

My questions:

1. Is this proof valid?
2. Are there faster way to make this proof?
3. Are there any "newbies-will-understand-this" proof? (a.k.a. not using Euler's number, trigonometry, and such))

Answer 1

1. The proof itself is correct, except that you didn't justify that $e^a > 0$. I think this is an important point. Those 10th graders probably aren't learning about the complex exponential, and certainly not the complex logarithm. Your proof shows that the complex exponential never reaches $0$ given that the real exponential never reaches $0$, but it seems to me that the latter is what you're really interested in. The solution @copper.hat gave in the comments shows this. It's also worth thinking about why $e^a$ is positive, as you claim in your proof.

2. The argument in the comments is very fast. If you want to make the argument you gave more concise, you can note that $|e^{ib}| = 1$. Indeed, $e^{ib} = \cos(b) + i \sin(b)$ so $|e^{ib}| = \sqrt{\cos(b)^2 + \sin(b)^2} = 1$ by the Pythagorean theorem. Hence, $|e^{a + bi}| = e^a$ which is nonzero.

3. As I remarked in (1.), I don't think a newbie who is unfamiliar with trigonometry should be learning about the complex exponential at all! The actual question is about the real exponential, and the proof in the comments works well there too. Also, I certainly don't know a way to explain that $e^x$ is nonzero without using Euler's number, since that's just $e^1$.