# Thread: e ^ i pi = -1? what?

1. ## e ^ i pi = -1? what?

im quite baffled by this question i was recently asked
so why does e^ (i*pi) = -1?

Any information is greatly appreciated!

2. Well it is known that $\displaystyle e^{i\theta} = \cos \theta +i\sin \theta$

Now make $\theta = \pi$ what do you get?

3. Originally Posted by lochnessmonster
im quite baffled by this question i was recently asked
so why does e^ (i*pi) = -1? Any information is greatly appreciated!
Using infinite series representations for $\cos(t)~\&~\sin(t)$
we can show that $\displaystyle e^{\mathif{i}t}=\cos(t)+\mathif{i}\sin(t)$.
So let $t=\pi$.

4. Originally Posted by lochnessmonster
im quite baffled by this question i was recently asked
so why does e^ (i*pi) = -1?

Any information is greatly appreciated!
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...

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

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

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
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...

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

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

Kind regards

$\chi$ $\sigma$
In fact, any complex number can be written as $\displaystyle z = r\cos{x} + i\,r\sin{x}$

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

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

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

$\displaystyle = iz$.

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

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

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

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

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

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

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

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

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

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

$\displaystyle r = A$.

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

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

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

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

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

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

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

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

6. Originally Posted by Prove It
... in fact any complex number can be written as $\displaystyle z = \cos{x} + i\sin{x}$...
... that may be the most 'revolutionary' idea of the 'new millennium'... I suggest however all students to forget that in their school tests...

Kind regards

$\chi$ $\sigma$

7. Originally Posted by chisigma
... that may be the most 'revolutionary' idea of the 'new millennium'... I suggest however all students to forget that in their school tests...

Kind regards

$\chi$ $\sigma$
I've realised that technically speaking that should read $\displaystyle z = r\cos{x} + i\,r\sin{x}$, but the proof is identical :P

8. Originally Posted by Prove It
I've realised that technically speaking that should read $\displaystyle z = r\cos{x} + i\,r\sin{x}$, but the proof is identical :P
Technically speaking we should not be leaving error in post just because they are convienient.

CB

9. 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 $\pi$ (which is the period).

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