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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?

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Are you sure that's $z^2+1$ in the denominator, not $z^2+i$? The obstruction to being analytic is vanishing of the denominator, and $z^2+1$ vanishes at $\pm i$. – Gerry Myerson Feb 24 '13 at 11:21
@GerryMyerson: Sorry, its $i$. But still how would you do it, we can tell it is not analytic when the denominator vanishes, but still we can't say anything about other points. How can you tell it satisfies CR eqns? – user23238 Feb 24 '13 at 12:31
As @julien writes, the quotient of analytic functions is analytic, away from zeroes of the denominator. – Gerry Myerson Feb 24 '13 at 21:52
up vote 1 down vote accepted

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 $$ \{z\in\mathbb{C}\;;\; z\neq \pm e^{i\pi/4}\}\cap \{z\in\mathbb{C}\;;\; \mbox{Im}z\neq 0 \;\mbox{or}\; \mbox{Re}z>-4\}. $$

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