Looking through some questions on Complex Analysis I have come across an interesting one. In general, these concepts are not too difficult (branch points/cuts, analytic functions and complex logarithm) but together in a single question this has thrown me a bit.
Question: Prove the function $\frac{Ln(z-1)}{z^2+i}$ is analytic in the cut plane with the cut along the negative half of the real axis up to and including $x=1$ and punctured at the points $z=\pm\frac{1}{\sqrt2}(1-i)$
The last part of the question, namely showing that the function has branch points at $z=\pm\frac{1}{\sqrt2}(1-i)$ is not hard at all. One simply shows that, by observation, $\frac{Ln(z-1)}{z^2+i}$ has branch points at $z=\pm\sqrt{-i}$ as these are the values of $z$ for which the function is not defined, namely $z^2+i=0$.
Now, we can rewrite $\pm\sqrt{-i}$ as $\pm\sqrt{(\frac{1}{2}-i-\frac{1}{2})}=\pm\sqrt{\frac{1-2i-1}{2}}=\pm\sqrt{\frac{1-2i+i^2}{2}}=\pm\frac{1}{\sqrt{2}}(1-i)$ thus showing that the function has branch points at these values. Since we are talking about whether or not a function is analytic, we would say the function is punctured at these points, and this part of the proof is finished.
Of course, ideally I would be doing this at the end but I don't know how to prove something with a complex logarithm is analytic. I gave it a small try but got lost quite quick. It went something like this:
To show this function is analytic we use the Cauchy-Riemann equations: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$
To do this we must rewrite the complex logarithm in terms of $x$ and $y$ i.e $Ln(z)=ln(r)+iArg(z)$ where $r=|z|$ and $Arg(z)=arctan(\frac{y}{x})$
Now I have $Ln(z-1)=Ln((x-1)^2+y^2)$ where $r=\sqrt{(x-1)^2+y^2}$ and $Argz=arctan(\frac{y}{x-1})$ so $Ln(z-1)=ln(\sqrt{x^2-2x+1+y^2})+i(arctan(\frac{y}{x-1}))$
Plugging into the CR-equations I get $$\frac{\partial}{\partial x}(ln\sqrt{(x-1)^2+y^2})=\frac{x-1}{(x-1)^2+y^2}=\frac{\partial}{\partial y}arctan(\frac{y}{x-1})=\frac{x-1}{y^2+(x-1)^2}$$ so this satisfies the first condition and I get
$$\frac{\partial}{\partial x}(arctan(\frac{y}{x-1}))=-\frac{y}{(x-1)^2+y^2}, \frac{\partial}{\partial u}(ln\sqrt{(x-1)^2+y^2})=\frac{y}{(x-1)^2+y^2}$$ which shouldn't be true as that means $\frac{\partial v}{\partial x}=\frac{\partial y}{\partial y}$ when it should be $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$
Can someone please look over this and explain if I have done something wrong to cause the Cauchy-Riemann equations not to be satisfied or is this a by-product of the function only being analytic in the cut plane etc. If this is the case, or if it isn't, can anyone please explain how I would go about showing that $\frac{Ln(z-1)}{z^2+i}$ is analytic in the cut plane with the cut along the negative half of the real axis up to and including $x=1$?
EDIT: I fixed the Cauchy-Riemann Equation part, as that was a silly arithmetic error on my part, but the rest about the cut plane and stuff still holds and I am still lost.
Cheers!