Here $z^{1/2}$ is defined with respect to the principal branch of arg
How do I find where $\sqrt{2-z} $ analytic? I know
$$\sqrt{2-z} = e^{\frac{1}{2}log_Iz} $$
and I know $log_Iz$ is differentiable everywhere for $\theta \in (-\pi , \pi)$
How do I find where the cut line of $\sqrt{2-z} $ is from this?
I know the 2 will change $r$ to $r > 2$, but I don't know how $\theta$ changes.
Best Answer
I think you have a fair idea of what $\sqrt{2-z}$ is in relation to $\sqrt{z}$.
The function $\sqrt{z}$ is analytic everywhere except along its branch cut, which is the positive real axis*
Then, $\sqrt{z-2}$ is analytic everywhere except along its branch cut, which is the real axis for $x\geq2$.
Finally, $\sqrt{2-z}$ is analytic everywhere except its branch cut, which is the real axis for $x \leq -2$.
*We have to cut at the positive real axis, because when we say $\displaystyle ({e^{iz}})^{\frac{1}{2}}$ we either mean $\displaystyle e^{\frac{iz}{2}}$ or $\displaystyle e^{\frac{iz}{2}+\pi i}$, and so the function kind of "resets" when you pass along the branch cut.
I think that normally, rotations and translations entire function are entire as well, however if there are some holes, poles, or branch cuts, you must translate them and rotate as appropriate.