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1. May 9, 2015

### TheDemx27

I've been using euler's formula now more than I have in the past, (using it for circuit analysis stuff), and so its been floating around in my head a bit more.

Say you have $e^{2πi}=1$ and you take the natural log of both sides.

$\log_e( e^{2πi})=\log_e(1)$
$2πi=0$
uhhhhh...

2. May 9, 2015

### axmls

The problem is that you are using the real natural logarithm, which is the inverse of the real exponential function $e^x$. You need to use the complex logarithm.

3. May 9, 2015

### TheDemx27

Ah, thankyou.

4. May 9, 2015

### certainly

The general equation is $e^{\pm 2ni\pi}=1$
The function $e^{ix}$ has periods of $2\pi$, just as the trigonometric functions have periods of $2\pi$
[i.e. although $sin(2\pi)=sin(0),\ 2\pi\neq0$]