How to prove $\operatorname{Log}(z) = \log(|z|)+i\arg(z)$. 
The value of the principal branch of the logarithm can be evaluated by the formula 
  \begin{align*}
\operatorname{Log}(z) = \log(|z|)+i\arg(z),
\end{align*}
  where $\arg(z) \in (-\pi,\pi)$ and $\log$ denotes the usual real logarithm.

So the strategy was to represent $z \in \mathbb C \setminus (-\infty,0]$ in polar form $z = re^{i\phi}$, where $r = |z|$ and $\phi = \arg(z)$. Then, if $\phi > 0$, let $\gamma = [1,r]+\gamma_2$, where $\gamma_2 = re^{it}$ with $t \in [0,\phi]$, and prove $\int_{[1,z]} \frac{1}{z} dz = \int_\gamma \frac{1}{z} dz$. I have computed the latter integral and I got exactly the wanted formula on the right handside. Then one can do similarly the case for $\phi < 0$ and we are done. 
My problem is, that I just don't see why $\int_{[1,z]} \frac{1}{z} dz = \int_\gamma \frac{1}{z} dz$ this holds.
I would really appreciate explanations here. Thanks.
 A: It follows from Cauchy's integral theorem that $\int_\gamma{dx\over z}$ will have the same value for any path $\gamma$ from $1$ to $z$ in $\mathbb{C}\setminus (-\infty,0]$. This is true since $\mathbb{C}\setminus (-\infty,0]$ is simply connected ("has no holes") and $f(z)={1\over z}$ is holomorphic in this domain. Cauchy's integral theorem is a deep result, but given this you are free to choose your $\gamma$ as you wish, and the particular $\gamma$ that you have chosen is the one which makes the computation simplest.
If you are interested in $\int_{[1,z]}{dz\over z}$, this can also be computed directly with elementary (and real) means, and will yield the same answer, but with a slightly more involved computation. (Note by the way that the notation is somewhat overloaded, in that $z$ denotes both the variable which runs along the interval, as well as the end point of the interval.) 
Assume that $z=x+iy$ with $y\neq 0$, since otherwise there is no real difference between the two integrals. Parametrize $[1,z]$ by $t\mapsto t\cdot z+(1-t)\cdot 1$ with $0\leq t\leq 1$. Then
$$\int_{[1,z]}{dz\over z}=\int_0^1{z-1 \over 1+t(z-1)}dt$$
Write the integrand of the last integral on standard form $a+ib$ for complex numbers by expanding the fraction by the complex conjugate of the denominator. This yields that the integral is equal to
$$\int_0^1 \biggl({(x-1)(1+t(x-1))+ty^2 \over (1+t(x-1))^2+t^2y^2} + i{y\over (1+t(x-1))^2+t^2y^2}\biggr)dt$$
The integral of the first term is equal to
$$ {1\over 2}\log\bigl((1+t(x-1))^2+t^2y^2\bigr)\Bigr|_0^1={1\over 2}\log(x^2+y^2)=\log(|z|)$$
and the integral of the second term is (with some work) shown to be equal to
$$ i\arctan\biggl({t(x^2+y^2-2x+1)+x-1 \over y}\biggr)\bigg|_0^1$$ $$=i\arctan\biggl({x^2+y^2-x\over y}\biggr)-i \arctan\biggl({x-1\over y}\biggr)=i\arctan {y\over x}=i \arg(z)$$
