Question

hi how to prove e^i(pi)=-1

Answer 1

\[\Large\begin{array}{llr}
&e^{i\pi}\\
=&\cos\pi+i\sin\pi&\mbox{(Euler's formula)}\\
=&-1+0i\\
=&-1
\end{array}\]

Answer 2

https://en.wikipedia.org/wiki/Euler's_formula

Answer 3

That kind of depends on if you believe euler's magical looking formula is true or not. I barely believe it myself, why should there be any relation between euler's number and the quadratic forms?

But if you expand out e^x, sinx, and cosx into infinite polynomials it makes it more obvious when you plug in i. Weird stuff.

Answer 4

\[\begin{align*}e^{ix}&=\sum_{n=0}^\infty \frac{(ix)^n}{n!}\\\\\\
&=1+ix+\frac{i^2x^2}{2!}+\frac{i^3x^3}{3!}+\frac{i^4x^4}{4!}+\frac{i^5x^5}{5!}+\frac{i^6x^6}{6!}+\cdots\\
&=\left(1+\frac{i^2x^2}{2!}+\frac{i^4x^4}{4!}+\frac{i^6x^6}{6!}+\cdots\right)+\left(ix+\frac{i^3x^3}{3!}+\frac{i^5x^5}{5!}+\cdots\right)\\
&=\left(1+\frac{i^2x^2}{2!}+\frac{i^4x^4}{4!}+\frac{i^6x^6}{6!}+\cdots\right)+i\left(x+\frac{i^2x^3}{3!}+\frac{i^4x^5}{5!}+\cdots\right)\\
&=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)\\\\\\
&=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{(2n)!}+i\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\\\\\\
&=\cos x+i\sin x
\end{align*}\]
So, when \(x=\pi\), you get \(e^{i\pi}=\cos \pi+i\sin \pi=-1+i\times 0=-1\).

Answer 5

A fun little extra:

\[e^x=e^{i(-ix)}=\cos(ix)-isin(ix)=\cosh(x)+\sinh(x)\]

Answer 6

@Kainui Wow I didn't know this thanks :D

Answer 7

never thought of mclaurin's thanks a loads, @SithangGiggles