# Can any one explain e^i^pi=-1?

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##### Newbie level 6
can any one explain e^i^pi=-1

#### v_c

Re: maths

with $i=\sqrt{-1}$, $e^{-i \pi}= \cos(\pi) + i \sin(\pi)=-1+0=-1$. this is called Euler's Identity. You can derive it by looking at the Taylor series expansion of $e^x$ and substitute $x=i \pi$. It is explained very well here

Euler's formula - Wikipedia, the free encyclopedia
(Make sure you enter the correct URL as the bit after the ' is not highlighted)
(look down the page where it says "Using Taylor Series")

I hope this helps you out.

Best regards,
v_c

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