$\pi = 0$! It can't be!

Question

Keep in mind: I'm still in high-school so forgive my poor maths. Also remember that I'm in HIGH-SCHOOL so nothing to complex

I like to mess around with equations and I find it quite fascinating the results I can somehow come up with. Recently, I found Euler's Identity and started to mess around with this. Somehow I can up with the following result: $π = 0.$ I thought to myself, as anyone would, how is this possible?

I then decided to show my proof to my maths teacher and my parents and neither one could show me where I went wrong so I was hoping someone here could.

This is my proof:

$$e^{iπ} + 1 = 0$$ $$e^{iπ} = -1$$ $$e^{2iπ} = 1$$ $$2iπ = \ln(1) = 0$$ $$4\times-1\timesπ = 0$$ $$-4π^2 = 0$$ $$0 = 4π^2$$ $$0 = 2π$$ $$0 = π$$

Edit

As I pointed out above, my maths knowledge isn't that advanced, compared to people on this site. Therefore the answer(s) I accept and/or upvote will be based on understandability as well as how well they answer the question.

Be careful with taking logarithms of complex numbers: https://en.wikipedia.org/wiki/Complex_logarithm#Problems_with_inverting_the_complex_exponential_function — Anna SdTC, Mar 20 '17 at 20:06
Oh, my dear, when we consider complex numbers, then it follows that$$\ln(1)=2\pi ik$$and more generally,$$\ln(z)=\ln|z|+i\operatorname{arg}(z)+2\pi ik$$Here is a good link to read: https://en.wikipedia.org/wiki/Complex_logarithm#Problems_with_inverting_the_complex_exponential_function — Simply Beautiful Art, Mar 20 '17 at 20:07
Since the exponential is not one-to-one on the complex numbers, it does not have a well-defined inverse. — Aweygan, Mar 20 '17 at 20:07
Well, do you know Euler's formula? It basically goes as follows:$$e^{\varphi i}=\cos(\varphi)+i\sin(\varphi)$$Thus, we have$$e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1=\cos(0)+i\sin(0)=e^0$$But $2\pi i\ne0$. We get an equality here due to the trig functions. — Simply Beautiful Art, Mar 20 '17 at 20:22
I find the 6 answers to be hilarious. Sorry for saying so, but I'm still laughing. I mean, Dah! How many times does this need to be asked and answered, and none the less, there are fish who bite the bait. Are any of the six fish who bit saying anything that none of the other fish who bit are saying?! — amWhy, Mar 20 '17 at 20:44
Hahahaha! That's good, @Jaideep! Some folks seem to forget that ! is not just punctuation, and especially not so on a math site! — amWhy, Mar 20 '17 at 20:55

score 2 · Answer 1 · answered Mar 20 '17 at 20:13

2

If $x$ and $y$ are real, then $e^x=e^y$ implies $x=y$.

But if $x$ and $y$ are complex, $e^x=e^y$ only implies that $$\frac{x-y}{2\pi i}\in\Bbb Z$$

Therefore, from $e^{2\pi i}=e^0$ we can't deduce that $2\pi i=0$.

answered Mar 20 '17 at 20:13

ajotatxe

66,849

score 0 · Answer 2 · answered Mar 20 '17 at 20:11

$e^z = e^w$ does not imply that $z = w$; the exponential function is not one-to-one on ${\mathbb C}$. Instead, it implies that $z - w = 2 \pi i n$ for some $n \in {\mathbb Z}$. So, the fourth equality of your 'proof' does not follow from the third equality.

score -1 · Answer 3 · answered Mar 20 '17 at 20:11

-1

The problem is simple. Note that if $0=\log(1)=\log(e^{2\pi I})=2\pi i$ than the logarithm is no mapping. Canonically one give a solution to this type of equations by restricting to the main branch of the complex logarithm.

answered Mar 20 '17 at 20:11

user427574

1,465

score -1 · Answer 4 · answered Mar 20 '17 at 20:14

The $log(z)$ or $ln(z)$ is a set not a value!

It denotes all $z' \in \mathbb{C}$ such that $e^{z'} = z$

What you want is the $Log(z)$ with $arg(Log(z)) \in [0,2\pi)$ or $(-\pi,\pi)$. Depends on how you define it.

As other said already it's a bit more difficult in $\mathbb{C}$.

score -1 · Answer 5 · answered Mar 20 '17 at 20:14

"2iπ=ln(1)=0"

In complex numbers $\ln$ is not single value but a multi-valued "function". So $\ln 1$ doesn't just equal $0$. It equals the entire set of $\{2k\pi i\}$ of which $0$ is the only real value in the set.

This is because $e^{a + bi} = e^a(\cos b + i\sin b)$ and trig functions are periodic so $z= e^{a + bi} = e^{a +bi + i2k\pi}$ is not a 1-1 function any more. So $\ln z = \{a + i(b + i2k\pi)| k \in \mathbb Z\}$ has multiple values.

So from there we can conclude only that $0 = i 2k\pi$ for some integer value of $k$. Which it does. ... id $k = 0$.

$\pi = 0$! It can't be!

Edit

5 Answers5