im quite baffled by this question i was recently asked

so why does e^ (i*pi) = -1?

Any information is greatly appreciated!

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- Apr 11th 2011, 02:07 PM #1

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- Apr 11th 2011, 02:11 PM #2

- Apr 11th 2011, 02:12 PM #3

- Apr 11th 2011, 11:26 PM #4
Let's suppose that Leonhard Euler has been a poet and not a mathematician, so that we ignore his formula. In that case the answer to Your question may derive from the solution of the differential equation...

$\displaystyle z^{'}= i\ z\ , \ z(0)=1$ (1)

The (1) is the trajectory of a point the velocity of which is its postion rotated by $\displaystyle \displaystyle \frac{pi}{2}$ so that we have an 'uniform circular move'. The solution of (1) is $\displaystyle z=e^{i\ t}$ so that the question is: where will be the point at the time $\displaystyle t= \pi$?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

- Apr 11th 2011, 11:56 PM #5
In fact, any complex number can be written as $\displaystyle \displaystyle z = r\cos{x} + i\,r\sin{x}$

so it is quite easy to see that $\displaystyle \displaystyle \frac{dz}{dx} = -r\sin{x} + i\,r\cos{x}$

$\displaystyle \displaystyle = i^2r\sin{x} + i\,r\cos{x}$

$\displaystyle \displaystyle = i(r\cos{x} + i\,r\sin{x})$

$\displaystyle \displaystyle = iz$.

Since $\displaystyle \displaystyle \frac{dz}{dx} = iz$

$\displaystyle \displaystyle \frac{1}{z}\,\frac{dz}{dx} = i$

$\displaystyle \displaystyle \int{\frac{1}{z}\,\frac{dz}{dx}\,dx} = \int{i\,dx}$

$\displaystyle \displaystyle \int{\frac{1}{z}\,dz} = ix + C_1$

$\displaystyle \displaystyle \log{|z|} + C_2 = ix + C_1$

$\displaystyle \displaystyle \log{|z|} = ix + C$ where $\displaystyle \displaystyle C = C_1 - C_2$

$\displaystyle \displaystyle |z| = e^{ix + C}$

$\displaystyle \displaystyle |z| = e^Ce^{ix}$

$\displaystyle \displaystyle z = Ae^{ix}$ where $\displaystyle \displaystyle A = \pm e^C$.

So $\displaystyle \displaystyle z = r\cos{x} + i\,r\sin{x} = Ae^{ix}$, and since we know that when $\displaystyle \displaystyle x = 0, z = r$

$\displaystyle \displaystyle r = A$.

Therefore $\displaystyle \displaystyle r\,e^{ix} = r\cos{x} + i\,r\sin{x}$.

Now by letting $\displaystyle \displaystyle x = \pi$ we have

$\displaystyle \displaystyle r\,e^{i\pi} = r\cos{\pi} + i\,r\sin{\pi}$

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

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

$\displaystyle \displaystyle e^{i\pi} = -1$

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

This is considered to be the most beautiful equation known, because it links the five fundamental constants of mathematics.

- Apr 12th 2011, 01:17 AM #6

- Apr 12th 2011, 01:55 AM #7

- Apr 12th 2011, 02:01 AM #8

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- Apr 12th 2011, 04:07 AM #9

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there are many ways of "deriving" this truth. some require you to have some knowledge of complex numbers, or calculus, upon which a sort of proof is based.

i, however, will not do this. instead, i will try to give you a sense of what e^z MEANS, when z is a complex number.

one can write a complex number as a two-part sum, one component in the "x" or "real" direction, and another in the "y" or "imaginary" direction. so we can write z as follows:

z = x + iy. now if we assume that exponents obey some of the rules we have come to expect in the real case, we ought to have:

e^z = e^(x+iy) = (e^x)(e^iy).

and so far, so good, and the e^x part behaves as expected, as x gets bigger, it gets bigger. as x becomes very negative, it tends toward 0.

but the overall effect is not to transform the plane into some inverted parabola-like shape, as you might expect. and this is because of the e^(iy) part.

if you hold x constant, and travel along the y-direction, e^(iy) traces out a circle in the complex plane. up-and-down, rather than being strictly increasing,

e^(iy) is periodic, like a trig function. so while on the left-hand side of the y-axis, e^z is rather flat, on the right-hand side it increases rapidly to the left in amplitude,

and undulates up and down as you travel either up, or down, in a delicate balance of interacting sines and cosines. along the vertical line x = 0,

e^(iy) is bending the y-axis into a circle of radius 1, starting anew every 2$\displaystyle \pi$ (which is the period).

and so it happens, that at y = $\displaystyle \pi$, we are half-way 'round a circle-cycle, and the value of the combined coordinates is -1 + 0i