Show that $F'(Z)=\frac{1}{Z}$ if $F(Z)=\log Z$ and $\frac{-\pi}{4}< \operatorname{arg}Z<\frac{7\pi}{4}$ Show that $F'(Z)=\frac{1}{Z}$ if $F(Z)=\log Z$ and $\frac{-\pi}{4}< \operatorname{arg}Z< \frac{7\pi}{4}$. Can I use definition of derivative here? What is the reason of adding condition on $\operatorname{arg}Z$? What will happen if the range of the argument is changed?
 A: The function $\log z$ is not continuous on any set containing a closed path around zero, so you need to restrict it to a subset of the plane that contains no such paths, for example, you "throw away" the ray from 0 to infinity directed $-\frac{\pi}{4}$ rad (which is the same as $\frac{7\pi}{4}$). So the reason for the restriction is making the function continuous and then, it's not hard to check that it'll be analytic too.
You can use the definition of the derivative once you know the function is analytic.
A: In the first place you should use your definition of $\log$. 
A first definition: One  chooses a suitable simply connected domain $\Omega\subset {\mathbb C}\setminus\{0\}$ containing $1$ and defines  $\log$ as
$$\log z:=\int_1^z{d\zeta \over\zeta}\ .$$
Then it is obvious that ${d\over dz}\log z={1\over z}$. 
Another way is the following: Choose a continuous branch ${\rm Arg}:\ \Omega\to{\mathbb R}\ $ of the polar angle  $\arg:\ {\mathbb R}^2\to{\mathbb R}/(2\pi{\mathbb Z})$ and put
$$\log(x+iy)\ :=\ {1\over2}\log(x^2+y^2) + i\,{\rm Arg}(x,y)\ .$$
Then check by means of the CR equations that $\log$ is in fact an analytical function of the complex variable $z:=x+iy$.
In any case, restrictions on $\arg z$ only serve to make $\log$ well defined and have no effect on its derivative, which is always the function $z\mapsto {1\over z}$.
