e^(i*pi) + 1 = 0
Few things in mathematics fascinate me as much as this simple, seemingly-trivial statement. Inside this one equation, we have all of the basic constants of mathematics (e, i, pi, 1, & 0), but we also incorporate trigonometry (admittedly, it's behind the scenes, but it's there). To understand this identity requires knowledge from several different areas of mathematics. Thus, fully understanding it sort of brings about a sense of fully understanding elementary mathematics. (And, for me personally, it doesn't hurt that my favorite mathematician, Leonhard Euler, figured out the mathematics behind the identity.)
So, how does the identity work? (I'm so glad I asked! [grin]) It turns out that e^(ix) can be evaluated as the sum of cos x and i sin x, that is,
e^ix = cos x + i sin x
(The explanation for this is far behind the scope of a simple blog post. If you want to find out why, consult a good mathematics textbook.)
In the case of the identity above, x = pi, so
e^(i*pi) = cos pi + i sin pi
cos pi = -1, and sin pi = 0; therefore, we have
e^(i*pi) = -1 + i*0 = -1
This result means that the first term in the identity is simply -1, making the result of its addition to 1 be 0.
That is how the identity works, in case you were curious.
2 comments:
An absolutely fundamental theorem to the understand of electrical engineering as well, since so much of AC circuit analysis requires an understanding of complex numbers and being able to convert between rectangular form and polar form.
You have a point there. I hadn't thought of its application to EE. Thanks for pointing that out!
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