# How to prove e^ix=cos x + i sin x

1. Aug 4, 2006

### HeilPhysicsPhysics

How to prove e^ix=cos x + i sin x

2. Aug 4, 2006

### d_leet

One way is to start with the taylor series for ex and then change x to ix and remembering that
i2 = -1, i3 = -i, and i4 = 1 you can rearrange the series and show that this is equal to the other side of that equation.

3. Aug 4, 2006

### Data

The proof depends on your definitions (for example, if you define $\cos{x}$ as $(e^{ix}+e^{-ix})/2$ and $\sin{x}$ as $(e^{ix} - e^{-ix})/(2i)$ then it's pretty easy!).

It's also clear if you start from the Taylor series of the three functions about the origin as d_leet suggested.

You can also start from the differential equation for e^x,

$$y^\prime = y, y(0)=1.$$

Write

$e^{ix} = f(x) + ig(x),$

(where f, g are real-valued)

so that $f(0) = 1$ and $g(0) = 0$

then you must have (by the chain rule)

$$ie^{ix} = f^\prime(x) + ig^\prime(x),$$

or, rearranging,

$$e^{ix} = \frac{1}{i}f^\prime(x) + g^\prime(x) = g^\prime(x) - if^\prime(x).$$

Since $e^0 = 1$ that means $g^\prime(0) = 1$ and $f^\prime(0) = 0$. Now differentiate again and you get

$$ie^{ix} = g^{\prime \prime}(x) - if^{\prime \prime}(x)$$

or

$$e^{ix} = -f^{\prime \prime}(x) - ig^{\prime \prime}(x).$$

Comparing the real and imaginary parts with our original $e^{ix} = f(x) + ig(x)$ you get two differential equations with initial conditions:

$$f^{\prime \prime}(x) = -f(x), f^{\prime}(0) = 0, f(0) = 1$$

and

$$g^{\prime \prime} (x) = -g(x), g^{\prime}(0) = 1, g(0) = 0.$$

It's easy to check that f(x) = cos(x) and g(x)= sin(x) satisfy these DEs, and by a theorem in the theory of DEs the solutions are unique.

Last edited: Aug 4, 2006
4. Aug 5, 2006

### Edgardo

Let $$z = \rm{cos}\theta + i \cdot \rm{sin}\theta$$

Then derive z:

$$\frac{\rm{d}z}{\rm{d}\theta} = - \rm{sin}\theta + i \cdot \rm{cos}\theta = -1 (\rm{sin}\theta - i \cdot \rm{cos}\theta) = i^2 (\rm{sin}\theta - i \cdot\rm{cos}\theta ) = i (i \cdot \rm{sin}\theta - i^2 \cdot \rm{cos}\theta) = i(\rm{cos}\theta + i \cdot \rm{sin}\theta) = i \cdot z$$

This is a differential equation:
$$\frac{\rm{d}z}{\rm{d}\theta} = i z$$
whose solution is

$$z = e^{i \theta}$$

Thus, $$\rm{cos}\theta + i \rm{sin}\theta = e^{i \theta}$$

Last edited: Aug 5, 2006
5. Aug 8, 2006

### eljose

Umm.. in fact i think that the "historical" development fo the formula came with euler when studying the "Harmonic (classical) oscilator"..

$$y'' + \omega ^2 y =0$$ trying y=exp(ax) then you get..

$$a^2 +\omega ^2 =0$$ so $$+ \sqrt (-1) \omega =a$$ (and the same

with a minus sign)

-On the other hand the long-known solution for the Harmonic movement was $$y= Asin (\omega t) +B Cos(\omega t)$$ then if you set the initial condition $$y(0)=0$$ you get the identity for sin. and using the well-known formula $$cos^2 (x) + sen^2 (x) =1$$ oh..if today math were so easier....i would have became famous years ago....

Last edited: Aug 8, 2006