1. ## complex logarithm

Let $z = x+iy$

I want to show that $\ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz$

$\ln(e^{iz}) = \ln|e^{iz}| + i \tex{Arg} (e^{iz})$

$= \ln (e^{-y}) + i \text{Arg} (e^{iz})$

$= -y + i \text{Arg} (e^{iz})$

EDIT: $\text{Arg} (e^{iz})$ is some angle in the interval $(-\pi, \pi]$. I don't understand how I can express that with use of the floor function.

2. Originally Posted by Random Variable
Let $z = x+iy$

I want to show that $\ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz$

$\ln(e^{iz}) = \ln|e^{iz}| + i \tex{Arg} (e^{iz})$

$= \ln (e^{-y}) + i \text{Arg} (e^{iz})$

$= -y + i \text{Arg} (e^{iz})$

EDIT: $\text{Arg} (e^{iz})$ is some angle in the interval $(-\pi, \pi]$. I don't understand how I can express that with use of the floor function.
Dear Random Variable,

Whatever the value of x; $\lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor$ is an integer. That is,

$\lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor=n\mbox{ where }n\in{Z}$

Consider, $\exp{\left( 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor+iz\right)}=\exp{(2\pi ni+iz)}=e^{2\pi ni}.e^{iz}=e^{iz}~;~(e^{2\pi ni}=1)$

Hence, $\ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz$