This problem is from Churchill and Brown. How do I prove that $f(z)=\frac{Log(z+4)}{z^2+i}$ is analytics everywhere except $\pm\frac{(1-i)}{\sqrt{2}}$ and on the portion $x \le -4$ of the real axis.
where Log is the prinicipal branch of the logarithmic fn.
I have expanded $Log(z+4)=\frac{1}{2}\ln((x+4)^2+y^2)+\arctan(y/x+4)$, and then reduced the whole fn to $u(x,y)+iv(x,y)$ form. However, the expression I am getting is quite complicated and would be hard to put put in the Cauchy-Riemann eqns.
Is there an easier method? Perhaps not converting into $x,y$ variables, but some direct theorem?
So $\log $ is the principal branch of the complex logarithm, which is defined and analytic on $\mathbb{C}\setminus (-\infty,0]$. I take that for granted. See here if you want details.
Therefore your numerator is analytic on $\mathbb{C}\setminus (-\infty,-4]$.
Now your denominator is analytic and nonzero on $\mathbb{C}$ minus the two roots of $z^2+i$.
Recall that the quotient of an analytic function by a nowhere zero analytic function is analytic.
So the domain of analyticity of your quotient is the intersection of these two open sets.
That is $$ \{z\in\mathbb{C}\;;\; z\neq \pm e^{i\pi/4}\}\cap \{z\in\mathbb{C}\;;\; \mbox{Im}z\neq 0 \;\mbox{or}\; \mbox{Re}z>-4\}. $$