First off, I am determining where the complex function $f(z)=Log(z+1)$ is analytic.
I am going by the definition that in order for the principal complex Logarithm to be analytic, the domain of $z$ need be such that $|z|>0$ and $-π<Θ<π$.
So for $$f(z)=Log(z+1)$$
$$=Log((x+1)+iy)$$
... where $$|z+1|>0$$
$$\sqrt {(x+1)^2+y^2}>0$$
(... which is true for all values of $z=x+iy$ except for $z=-1+i0=-1$.)
... and... $$-π<Θ=tan^{-1}(\frac{y}{x+1})<π$$
... my domain is simply $\{z=x+iy | x>-1\}$.
My reasoning being that it is unnecessary to state the "$y$ cannot equal $0$ while $x$ equals $-1$" part because $x$ cannot equal $-1$ to begin with as it violates the definition I stated at the top, not to mention in this case it causes division by zero in regards to my $Θ$. So am I right? I am assigned Churchill's book for this class and it is not too helpful, so I apologize if I am missing something very obvious.
P.S.: If anybody has the time for one more little question, why can't $$Θ=π$$ in the context of $f(z)=Log(z)$ such as it is usually allowed to throughout the rest of complex analysis.
How did you get $\{z=x+iy:x>0\}$? It is best to think geometrically. $Log (z)$ is analytic on $\mathbb C \setminus (-\infty, 0]$. To make the function analytic you have to remove all non-positive real numbers from the complex plane. To see where $Log (z+1)$ ia anlytic you simply have to choose $z$ such that $z+1 \notin (-\infty, 0]$ which means $z \notin (-\infty, -1]$. So the answer is $\mathbb C \setminus (-\infty, -1]$. For your second question you allow $\pi$ to be the argument of $z$ is the definition of Log but it becomes discontinuous at points with argument is $\pi$.