Is Euler's Formula Valid When Using a Non-Power Series Method?

[daster]

I was messing around and decided to prove Euler's 'formula' using a method that doesn't involve power series. Here's how I did it:

$$z=\cos\theta + i \sin\theta$$

$$\frac{dz}{d\theta} = -\sin\theta + i\cos\theta$$

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

$$-i\int \frac{1}{z} \; dz = \int \; d\theta$$

$$-i \log|z| = \theta + C$$

When [itex]\theta=0[/tex], $z=1$, so $C=0$. Now:

$$\log|z| = \frac{-\theta}{i}$$

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

 

[hypermorphism]

I was messing around and decided to prove Euler's 'formula' using a method that doesn't involve power series. Here's how I did it:

$$z=\cos\theta + i \sin\theta$$

$$\frac{dz}{d\theta} = -\sin\theta + i\cos\theta$$

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

$$-i\int \frac{1}{z} \; dz = \int \; d\theta$$

$$-i \log|z| = \theta + C$$

When [itex]\theta=0[/tex], $z=1$, so $C=0$. Now:

$$\log|z| = \frac{-\theta}{i}$$

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

Nice work. Calculate 1/i in terms of a+ib, where a and b are real.

 

[jamesrc]

Nothing as far as I can tell; you just didn't finish.

$$\frac{-\theta}i = \frac{-\theta i}{i^2} = \frac{-i\theta}{-1} = i\theta$$

ETA: Oops, too slow on the draw.

 

[daster]

Nice! I was too busy checking my first steps to notice that I had to simplify my answer!

Thanks.

 

[matt grime]

There are, though, several things you have sidestepped when doing that integral of 1/z, such as the fact that log is an many valued function, and why when you exp log of |z| you don't get |z|, for instance. I mean, in

log|z| = \theta/i

the lhs is strictly real and the rhs is strictly imaginary.