Is $e^{z}$ a one to one function?

Question

Let $z \in \mathbb{C}$ , $z = x+iy ; x,y \in \mathbb{R}$

$e^{z} = e^{x+iy}$ and that will become $e^{x}(\cos({y})+i\sin({y}))$

but then I know cosine and sine are periodic, both trigonometric functions aren't one-to-one function at all.

Therefore complex exponential isn't one to one function.

Is this correctly done?

You need only find a counter example. Consider $y_1=0$ and $y_2=2\pi$, for instance. — user170231, Aug 18 '20 at 19:56
@coffeemath No, I mean to consider two different $y_i$ as the imaginary parts of some complex number $z$. — user170231, Aug 18 '20 at 20:17
Does this solution answer your question? https://math.stackexchange.com/a/21195/798113 — , Aug 18 '20 at 20:33

score 2 · Answer 1 · answered Aug 18 '20 at 20:08

2

No. If it was injective then $\exp(2\pi xi)$ would be also injective but the latter function takes value 1 infinitely many times by the Euler formula.

answered Aug 18 '20 at 20:08

markvs

19,653

so did I do it correctly? – EM4 Aug 18 '20 at 20:10
1

I just looked at your post. Yes it is correct. – markvs Aug 18 '20 at 20:11
Now I am thinking about it what if cosine and sine are restricted, then it will be one to one. If it isn't, it isn't one to one. – EM4 Aug 18 '20 at 20:15
What does it mean "restricted cos"? – markvs Aug 18 '20 at 20:16
$\exp z$ is locally injective, that is every point has a neighborhood where the function is injective. – markvs Aug 18 '20 at 22:31
what you mean locally injective? Does the multiple of values of domain have the same output in the range. – EM4 Aug 18 '20 at 23:04
I gave the definition of locally injective. – markvs Aug 18 '20 at 23:05

Is $e^{z}$ a one to one function?

1 Answers1