I am trying to determine where $f(z)=(z-i)\log(z^2+1)$ is continuous.
My attempt:
Well, $z-i$ is a complex polynomial, so it is continuous in all of $\mathbb{C}$. Hence, the continuity of $f(z)$ is dependent on $\log(z^2+1)$. I know that $\log(w)$ is continuous on $\mathbb{C}$\ $(-\infty,0]$, so i figured that $\log(z^2+1)$ is continuous when $z^2+1>0$. So, \begin{align} z^2+1&>0 \\ x^2-y^2+2xyi&>-1 \\ \end{align} Equating real and imaginary, $$x^2-y^2>-1 \ \ \ \ \text{and} \ \ \ \ xy=0$$ There are two cases to consider.
Case 1: $$x^2-y^2>-1 \ ,\ x=0$$ this yields $-1<y<1$.
Case 2: $$x^2-y^2>-1 \ ,\ y=0$$ this yields the empty set.
But the answer is for $y\geq 1, y\leq -1$. Where have I gone wrong?
The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1\notin(-\infty,0]$ (you will have to give a special look to $z=i$.)