MHB What is the Branch Cut for log(z)?

[Amer]

What is the Branch cut for the log(z) ?

Correct me if I am wrong.
I know that the function $f(z) = e^z$ , is periodic function with period $2 \pi $
so to define the function $\log(z) $ we have to restrict the domain of $e^z$
for example taking the points $D : z \in \mathbb{C} $ such that $-\pi < arg (z) < \pi $
$e^z $ is one to one in D
Now define the function $g(z)= \log (z) $ which has the range D
The branch cut of $\log (z) $ describe the range not the domain of log ?

Thanks

 

[chisigma]

What is the Branch cut for the log(z) ?

Correct me if I am wrong.
I know that the function $f(z) = e^z$ , is periodic function with period $2 \pi $
so to define the function $\log(z) $ we have to restrict the domain of $e^z$
for example taking the points $D : z \in \mathbb{C} $ such that $-\pi < arg (z) < \pi $
$e^z $ is one to one in D
Now define the function $g(z)= \log (z) $ which has the range D
The branch cut of $\log (z) $ describe the range not the domain of log ?

Thanks

The branch point of a multivalued function is a point in the complex plane from which several branchs of the function depart. The function f(z) = ln z has only one branch point in z=0...

Kind regards$\chi$ $\sigma$

 

[alyafey22]

$w=e^z$ then if $-\pi <Im(z) < \pi $ the function $w$ is one to one. Now assume the following

$$z=e^w$$ and we want to find $w$ satisfiying the property then $|z|=e^u$ implies that $u = \ln |z|$ and $v=arg(e^z)=arg(z) \in (-\pi ,\pi ) $. Hence we have

$$w=\log(z) = \ln|z|+iarg(z) \,\,\,;\, arg(z) \in (-\pi , \pi ) $$

Hence defining the principle logarithm with branch cut on the negative real axis.