How can I prove that $\ln z=\ln|z|+i\arg(z)$? 
How to prove:
$$\ln z = \ln|z| + i \arg z$$

How to prove this equation? Maybe  this is an easy question.Please don't give this question a down arrow as for I am a new man in complex analysis.I need your help.
 A: Let  $$z=re^{i\theta}$$
$$|z|=|re^{i\theta}|=|r|\cdot|e^{i\theta}|=r$$
$$\implies \ln(z)=\ln(r)+\ln(e^{i\theta})=\ln|z|+i\theta=\ln|z|+i\operatorname{arg}(z)$$
A: A standard definition of complex logarithm is:

Complex logarithm of $z$ denoted by $\ln{z}$ is the complex number $a$
  that satisfies $e^a=z$ or it's the inverse of the function $e^z$

By Euler's formula we know that any complex number $z$ can be represented as $z=|z|e^{i\theta}$ where $|z|$ represents it's modulus (distance of the complex number $z$ from the origin) and $\theta$ is the argument ( the angle the line joining $z$ and origin makes with the real axis). 
Now, for the sake of the $z$ having a unique representation as $e^a$ we need to make the argument of  $z$ is unique. How can this be done?. We can restrict the argument $\theta$ to the interval $\left(-\pi,\pi\right]$. And we call this argument as 'principal argument'.
With the knowledge aquired from the above paragraphs $$z=|z|e^{i\theta}=e^{\ln{|z|}}\cdot {e^{i\theta}}$$
$$z=e^{\ln{z}+i\theta}$$
$$\Rightarrow \ln{z}=\ln{|z|}+i\theta$$
or
$$\Rightarrow \ln{z}=\ln{|z|}+i\arg{z}$$
Where $\arg{z}$ is the principal argument.
But, actually complex logarithm is a multivariate function . 
$$\Rightarrow \ln{z}=\ln{|z|}+i\left(\arg{z}+2n\pi\right) \space \quad n\in \mathbb{I}$$
.The proof of the above is similar without the existence of the condition $\theta$ is restricted.
A: You start by writing the complex number $z$ in polar form: $z = r\!\operatorname{e}^{i\theta}$, where $r$ is the modulus of $z$ and $\theta$ is the argument of $z$. Then use the law that $\ln  (ab) = \ln a + \ln b$.
\begin{array}{ccc}
z &=& r\!\operatorname{e}^{i\theta} \\
\ln z &=& \ln\left(r\!\operatorname{e}^{i\theta}\right) \\
&=& \ln r + \ln \left(\operatorname{e}^{i\theta} \right) \\
&=& \ln r + i\theta
\end{array}
Since $r = |z|$ and $\theta = \arg z$ we have $\ln z = \ln|z| + i \arg z$. Obviously, $\theta$ is only well-defined up to multiples of $2\pi$. For example, $\arg 1 = \ldots,-2\pi,0,2\pi,\ldots$
A: Hint: what happens when you evaluate
$$
e^{\ln z}
$$
and
$$
e^{\ln |z|+i\arg z}
$$
