1. ## complex logarythm function

Hey
let z1,z2...,zn be complex numbers in the right plane such that their product is also in the right plane.prove that:

Log(z1*z2*...*zn)=Log(z1)+Log(z2)+...Log(zn)

where Log(z) denotes the principal branch of the logarythm function.
Recall that if \displaystyle \begin{align*} z = r\,\mathrm{e}^{\mathrm{i}\,\theta } \end{align*} then \displaystyle \begin{align*} \log{ \left( z \right) } = \log{ \left( r \right) } + \mathrm{i}\,\theta \end{align*}, so
\displaystyle \begin{align*} \log{ \left( z_1 \right) } + \log{\left( z_2 \right) } + \log{ \left( z_3 \right) } + \dots + \log{ \left( z_n \right) } &= \log{ \left( r_1 \right) } + \mathrm{i}\,\theta_1 + \log{ \left( r_2 \right) } + \mathrm{i}\,\theta_2 + \log{ \left( r_3 \right) } + \mathrm{i}\,\theta_3 + \dots + \log{ \left( r_n \right) } + \mathrm{i}\,\theta_n \\ &= \log{ \left( r_ 1 \right) } + \log{ \left( r_2 \right) } + \log{ \left( r_3 \right) } + \dots + \log{ \left( r_n \right) } + \mathrm{i}\,\theta_1 + \mathrm{i}\,\theta_2 + \mathrm{i}\,\theta_3 + \dots + \mathrm{i}\,\theta_n \\ &= \log{ \left( r_1 \cdot r_2 \cdot r_3 \cdot \dots \cdot r_n \right) } + \mathrm{i}\,\left( \theta_1 + \theta_2 + \theta_3 + \dots + \theta_n \right) \\ &= \log{ \left( z_1 \cdot z_2 \cdot z_3 \cdot \dots \cdot z_n \right) } \end{align*}