I know that it’s true but why? I’ve never grasped this topic.
In: Mathematics
e^(i π) = -1 is a particular result from a more useful, more general relationship:
e^(i t) = cos(t) + i sin(t).
This equation can be derived from the Taylor series of each function (Calculus incoming! You have been warned):
e^x = 1/0! + x/1! + x^(2)/2! + x^(3)/3! + x^(4)/4! + x^(5)/5! + …
= Sum(x^(n)/n!, for n=0 to infinity)
sin(x) = x/1! – x^(3)/3! + x^(5)/5! – …
= Sum((-1)^(n)x^(2n+1)/(2n+1)!, for n=0 to infinity)
sin(x) = 1/0! – x^(2)/2! + x^(4)/4! – …
= Sum((-1)^(n)x^(2n)/(2n)!, for n=0 to infinity)
Now, if we look at what e^(x) does with imaginary numbers, we get
e^(i x) = 1/0! + (ix)/1! + (ix)^(2)/2! + (ix)^(3)/3! + (ix)^(4)/4! + (ix)^(5)/5! + …
= 1/0! + ix/1! – x^(2)/2! – ix^(3)/3! + x^(4)/4! + ix^(5)/5! + …
If we separate out the real parts from the imaginary parts (We’ll skip the proof about absolute convergence. Suffice to say we can rearrange terms as we like), we get
e^(i x) = (1/0! – x^(2)/2! + x^(4)/4! + …) + i(x/1! – x^(3)/3! + x^(5)/5! + …)
= Sum((-1)^(n)x^(2n+1)/(2n+1)!) + i Sum((-1)^(n)x^(2n)/(2n)!)
= cos(x) + i sin(x)
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If you haven’t taken calculus or gotten to infinite series yet, sorry if that went over your head. I’m not aware of any way to prove the identity without it. The key takeaway, though, is that
e^(i t) = cos(t) + i sin(t)
and when you plug π into that, you get
e^(i π) = cos(π) + i sin(π)
= -1 + i 0
= -1.
Likewise if you plug in 3π, 5π, 7π, etc.
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