Why is $e^{i\pi}= -1$

Question

Why does $ e^{i\pi} = -1 $ ?

I know that this form can be used to for instance act on a bloch's sphere (quantum mechanics) using it as $ e^{i\pi/4} $ will do a $ \frac{\pi}{4} $ rotation on the $x-y$ plane

But I found no answer on the internet.

[EDIT]: Moreover the greats answers, I recommend this really well made video for thoses who are not familiary of the Euler's identity.

[EDIT TO CLARIFY]:
Learning the basics of quantum information, I had to use this matrix: $ \begin{pmatrix} 1 && 0 \\\ 0 & e^{i\pi/4} \end{pmatrix} $ which is a gate named $ T(\psi) $ where $ \psi $ is a simple qubit as $ \begin{pmatrix} \alpha \\\ \beta \end{pmatrix} $.

$ \alpha $ and $ \beta $ are just chances to get the classical qubit's state there, for instance $ \psi = \begin{pmatrix} \alpha \\\ \beta \end{pmatrix} $ gets $ \alpha^2 $ chances to be (as classical state) $ \begin{pmatrix} 1 \\\ 0 \end{pmatrix} $ (denoted $ |0\rangle $) and $ \beta^2 $ to be $ \begin{pmatrix} 0 \\\ 1 \end{pmatrix} $ (denoted $ |1\rangle $) (both $ \alpha $ and $ \beta $ can be non null, this is known as superposition).

To visualize a qubit state we can use the Bloch Sphere
The Bloch sphere

This depends entirely on what you mean when you write $e^{i\pi}$. What does that expression mean to you? Depending on your answer to that question, this video has a nice explanation in it. — Arthur, Mar 05 '18 at 15:32
Also see https://math.stackexchange.com/questions/2093969/explanation-of-eulers-identity and https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi — David K, Mar 05 '18 at 15:48
Thank you all, sorry for the dumb question, I was not aware of the Euler's identity, this post helped me to get the concept introduced, i'll follow all your kindly pointed resources for a deeper understanding :) — Alexandre Daubricourt, Mar 05 '18 at 19:56
Possible duplicate of Identities derived from Euler's Identity — hardmath, Mar 05 '18 at 21:14

score 2 · Accepted Answer · answered Mar 05 '18 at 15:37

2

$$e^{ix}=1+(ix)+\dfrac{(ix)^2}{2!}+\dfrac{(ix)^3}{3!}+\dfrac{(ix)^4}{4!}...=1+ix+\dfrac{i^2x^2}{2!}+\dfrac{i^3x^3}{3!}+\dfrac{i^4x^4}{4!}...$$

$$=1+ix-\dfrac{x^2}{2!}-i\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...=\left(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}...\right)+i\left(x-\dfrac{x^3}{3!}...\right) = \cos x+i\sin x$$

So:$$e^{ix}=\cos x+i\sin x$$

And you know the rest if $x=\pi$

answered Mar 05 '18 at 15:37

Thank you, I would like to add a link of a video that I liked who treat this equation as you did – Alexandre Daubricourt Mar 05 '18 at 21:27
@AlexandreDaubricourt You're welcome! – Mar 05 '18 at 21:28

score 1 · Answer 2 · answered Mar 05 '18 at 15:33

If $z \in \mathbb{C}$, $e^z$ is defined to be the power series $$\sum_{n=0}^\infty \frac{z^n}{n!}$$ This, and the power series for $\sin x$ and $\cos x$ lead to the identity $$e^{i \theta} = \cos(\theta) + i \sin(\theta)$$ from which $e^{i \pi} = -1$ follows.

score 0 · Answer 3 · answered Mar 05 '18 at 15:31

0

HINT: $$\exp(ix)=\cos(x)+i\sin(x)$$

answered Mar 05 '18 at 15:31

Dr. Sonnhard Graubner

95,283

score 0 · Answer 4 · answered Mar 05 '18 at 15:33

0

We have Euler's Identity $e^{i\theta}=\cos \theta + i \sin\theta$.

So, for $\theta =\pi$ we get $e^{\pi i}=-1$

answered Mar 05 '18 at 15:33

Sujit Bhattacharyya

3,030

Why is $e^{i\pi}= -1$

4 Answers4