A Mathematician Went To A Bar

Let $z= \cos (x) + i \sin (x)$ , then
$\dfrac{dz}{dx} = - \sin (x) + i \cos(x) = i^2 \sin (x) + i \cos(x) = i (\cos (x) + i \sin(x)) = i (z)$ .
$\dfrac{dz}{z} = i \, dx$ .

Integrating both sides,
$\ln (z) = ix + C$
$e^{\ln (z)} = e^{ix + C}$
$z = e^{ix + C}$
$e^{ix + C} = \cos (x) + i \sin(x)$ .

Substituting $x= 0$ ,
$e^{i\times 0 + C} = 1$
$e^{0 + C} = e^0$
$\therefore C= 0$

We get the Euler's equation,
$e^{ix} = \cos (x) + i \sin(x)$
When $x = \pi$ ,
$e^{i \pi} = \cos(\pi) + i \sin(\pi) = -1$
$\therefore -e^{i \pi} = -(-1) = \boxed 1$ .

I don't think you even need the math to solve it.. The word "pint" says that the answer is 1 !!

According to Euler's identity $e^{i\pi}+1 =0$ .

Euler's identity is a special case of Euler's formula from complex analysis, which states that for any real number $x$ ,

$e^{ix} = \cos x + i\sin x$ where the inputs of the trigonometric functions sine and cosine are given in radians.

In particular, when $x = \pi$ , or one half-turn $(180°)$ around a circle:

$e^{i \pi} = \cos \pi + i\sin \pi$ . Since

$\cos \pi = -1$ , and

$\sin \pi = 0$ , it follows that

$e^{i \pi} = -1 + 0 i$ , which yields Euler's identity:

$e^{i \pi} +1 = 0$ .

Well

$\begin{aligned}-e^{i\pi} &=-(-1) \\& \therefore 1\end{aligned}$ .

$\large\text{FIN!!}$