continuous square root complex function

Question

I have a series of problems in which I have to find a continuous function that agrees with certain conditions, but I fail to see how to even start thinking about such functions. I will provide an example: for $D=\mathbb{C}\backslash\left\{ \left(2+i\right)t\,\,\,:\,\,\,-1\leq t\leq1\right\} $ I need $f:D\to \mathbb C$ s.t: $$ \left(f\left(z\right)\right)^{2}=\left(z-\left(2+i\right)\right)\left(z+\left(2+i\right)\right)$$ and $$f\left(-2+i\right)=2-2i$$ My problem is with the discontinuity of $Arg(z)$ around $z=-1$, so complex numbers just above the $Re(z)<0$ line go from $\approx \pi$ to $\pi/2$, but complex number below go from $\approx -\pi$ to $-\pi/2$ and $e^{i\pi/2} \neq e^{-i\pi/2}$. How can I resolve this issue with these square functions? I don't even see how the definition of $D$ helps me.

please excuse me if this question is too "basic" for this tagging, I found nowhere else to get a clue how to proceed. I don't need or want you to fully solve this, just a clue please

This is much better than your earlier question - here it's clear why the discontinuity of arg matters. In fact the problem you point out means it's not possible to give a continuous $\sqrt{ \left(z-\left(2+i\right)\right) }$ in $D$, nor a continuous $\sqrt{\left(z+\left(2+i\right)\right)}$. But the two problems cancel because $(-1)^2=1$; you can piece together those discontinuous square roots of the factors to get a continuous $\sqrt{\left(z-\left(2+i\right)\right)\left(z+\left(2+i\right)\right)}$. — David C. Ullrich, Apr 09 '21 at 15:20
First of all thank you! Isn't it a problem though that the discontinuity of each of them happens at different points? in $z=i$ for the $\sqrt{z-(2+i)}$ for example, while the other function doesn't have a discontinuity at this point so nothing cancels. — Darkenin, Apr 09 '21 at 16:11
That's a problem or not, depending on what "branches" you choose for the two partial square roots... — David C. Ullrich, Apr 09 '21 at 17:03
It might be extremely foolish, but can I generally succeed in doing so by defining $Arg\left(\left(z-(2+i)\right)\cdot e^{-iarctan\left(\frac{1}{2}\right)}\right)$ if I wish the discontinuity be where I want it to in this case? Is that what you meant? — Darkenin, Apr 09 '21 at 20:07

score 1 · Answer 1 · edited May 28 '21 at 10:59

Exercise What you want to prove is equivalent to the following

Prop. Let $D=\Bbb C\setminus[-1,1]$. There exists $f\in H(D)$ with $f(z)^2=z^2-1=(z-1)(z+1)$. $\newcommand\Arg{\text{Arg}}$

The reformulation means that comments about placing the discontinuities in the right place resolve to just "always use the principal value argument $\Arg$ for the argument".

Note that $\Arg(z-1)+\Arg(z+1)$ is continuous at every point of $D$ except for $B=(-\infty,-1)$. On the other hand, if $x\in B$ then $\Arg(z-1)+\Arg(z+1)$ has a jump of $$2\pi+2\pi=4\pi$$ at $x$; hence $$e^{\frac i2(\Arg(z-1)+\Arg(z+1))}$$is continuous at $x$. So we can use $$f(z)=|z^2-1|^{1/2}e^{\frac i2(\Arg(z-1)+\Arg(z+1))}$$to get a continuous square root. (You can easily factor that to show that it is in fact of the form $\sqrt{z-1}\sqrt{z+1}$, as I suggested should be possible in the comments.)

Now we show as an exercise that a continuous square root must be holomorphic. Or apply this.

continuous square root complex function

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