# Why does the “typical” branch of $\text{log}\ z$ have the property $-\pi \lt Im(\text{log}\ z) = Arg\ z \lt \pi$?

I am reading the definition of the "typical" analytic branch of $\text{log}\ z$, which my book gives to be:

$$f(z) = \int_{1}^{z}{\frac{d\zeta}{\zeta}}$$

My book states this analytic branch on the entire complex plane minus the non-positive real axis: $x \leq 0$. It also states the following is true:

$$-\pi \lt \text{Im(log }z\text{)} = \text{Arg }z \lt \pi$$

The justification for this is: "This can be see by integrating from 1 to |z| and from |z| to z"

I don't understand how this justifies the inequality. Can someone explain? Thanks.

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If you express a complex number in polar coordinates, i.e. $z=re^{i\theta}$ and $\theta$ equal $\pi$, where does that point lie in the plane? –  user69810 Apr 18 '13 at 5:38

The first thing to realize might be that the RHS of the identity "defining" $f(z)$ does not specify a unique complex number. For example, the path starting at $1$ and running along the unit circle in the ENWS direction until it reaches $i$ and the path starting at $1$ and running along the unit circle in the ESWN direction until it reaches $i$ give two different values. Hence one cannot allow every path.
The miracle is that forbidding the paths to meet the ray $\{z\in\mathbb C\mid\Re(z)\leqslant0,\Im(z)=0\}$, is enough to determine uniquely the RHS. Note that there are plenty of other ways to determine uniquely the RHS, one can forbid any ray starting from $0$ or even any injective path starting at $0$ and going to infinity (for example a spiral such as $s:\mathbb R\to\mathbb C$, $t\to s(t)=\mathrm e^{(1+\mathrm i)t}$...).
Indeed, the path from $1$ to $|z|$ is parametrized by $\zeta=t$, $t$ real going from $1$ to $r=|z|$, hence the corresponding integral $g(z)=\int_1^r\frac{\mathrm dt}t$ is real. The second part, from $|z|$ to $z$, is parametrized by $\zeta=r\mathrm e^{it}$, $r=|z|$, $t$ going from $0$ to $\theta=\mathrm{Arg}(z)$. Then $\mathrm d\zeta=ir\mathrm e^{it}\mathrm dt=i\zeta\mathrm dt$ hence this part of the integral reads $h(z)=\int_0^\theta i\mathrm dt=i\theta$. Finally, $f(z)=g(z)+h(z)$ with $g(z)$ purely real and $h(z)$ purely imaginary, hence $\Re f(z)=g(z)=\log r=\log|z|$ and $\Im f(z)=\Im h(z)=\theta=\mathrm{Arg}(z)$.