The Euler formula how was it developed?

[Skynt]

I'm reading a book called The Road to Reality by Roger Penrose, and I'm on the chapter for complex logarithms. What I don't understand is how the identity e^{\theta i} = cos \theta + i sin \theta is found through the use of complex logarithms. I also don't understand how if w = e^z, z = log r + i\theta if w is in [r, \theta] form.

Basically, he explains in this book that e was chosen as a base in the general form w = b^z because it reduces the ambiguity of b^z.

I guess I'm just not seeing how the complex logarithm comes into play for the Euler formula being developed.

EDIT: Sorry about the format of the post, I was just trying latex. I hope someone can clear this up for me, because I feel it's something obvious I'm not understanding.

 

[mathman]

Off hand I don't know how Euler first discovered his identity, but I don't think natural logs were involved. However, the easiest way to prove the identity is to expand both sides into power series and see that these series are identical.

 

[maze]

First you have to think about what it means to write e^{z} for complex z. One way is via power series as already mentioned. Another is in terms if the differential equation \frac{d}{dz}e^z = e^z, e⁰=1 (e^z is its own derivative), which I imagine is how the concept arose historically.

If you express e^{i z} = u(z) + iv(z) for real u and v and use the differential equation definition, then simple computation will show that e^{i z} has length 1, its derivative has length 1, and (\frac{d}{dx}e^x)(0)=i, so the curve z->e^{i z} must trace out the unit circle in the u-v plane counterclockwise at unit-speed. (sin,cosine) also traces out the unit circle at unit speed, so Euler's formula follows.

 

[qntty]

Maybe this is the proof you mean (I saw this somewhere on mathworld I think)?
let z=\cos{\theta}+i\sin{\theta}
then \text{d}z = -\sin{\theta} + i\cos{\theta} \, \text{d}\theta= i(\cos{\theta}+i\sin{\theta}) \, \text{d}\theta

The integral
\int{\frac{\text{d}z}{z}}
can be evaluated in two ways. First you can say
\int{\frac{\text{d}z}{z}} = \log{z} = \log{(\cos{\theta}+i\sin{\theta})}
On the other hand you can substitute in dz and z to get
\int{\frac{i(\cos{\theta}+i\sin{\theta})}{\cos{\theta}+i\sin{\theta}}} \, \text{d}\theta =\int{i \, \text{d}\theta
so
\log{(\cos{\theta}+i\sin{\theta})} = i\theta

e^{i\theta} = \cos{\theta} + i\sin{\theta}

 

[ExactlySolved]

I believe I have seen the page of Euler's introductio in analysin infinitorum where he gives the proof in terms of power series.