I still don't get how to work with branches. I understand that it is a way to define continous multivalued functions, but how to apply it to an specific problem I still don't know how to do it.

Here is an exercise from Gamelin's Complex analysis (p53):

Consider the branch of $f(z)=\sqrt{z(1-z)}$ on $\mathbb{C}-[0,1]$ that has positive imaginary part at $z=2$. What is $f'(z)$? Be sure to specify the branch of the expression for $f'(z)$?

I have several questions:

  1. Regarding the function $f(z)=\sqrt{z(1-z)}$. I know that this function have "two special points", at $z=0$ and $z=1$, meaning that if I make a turn around one of these points, I get an extra phase. Question: why I mus exclude $[0,1]$, and not just the points $0$ and $1$?

  2. There is no problem in getting the derivative (it's just routine calculus), but what does it mean "... that has positive imaginary part at $z=2$"?

  3. Finally, how do I get the branch expression for $f'(z)$ and why to choose that branch?

  1. As you say, if you move around a small circle centred at $0$ or $1$ trying to keep $f(z)$ continuous, you end up with a value that is $-$ what you started with. So there is no version of this function that is continuous on such a circle (or more generally, on any closed curve that has one of $0$ and $1$ inside and the other outside). Any open set that doesn't include such a closed curve must exclude a curve joining $0$ to $1$ or two curves joining $0$ to complex$\infty$ and $1$ to complex $\infty$.
  2. There are two possible values for $f(2) = \sqrt{-2}$, namely $\sqrt{2}i$ and $-\sqrt{2}i$. You want $+\sqrt{2}i$. Once you have specified that, the requirement of continuity on $\mathbb C - [0,1]$ determines the value of $f$ at every other point.
  3. Since $f(z)^2 = z (1-z)$, $2 f(z) f'(z) = 1 - 2 z$, i.e. $$ f'(z) = \dfrac{1-2z}{2 f(z)} $$
  • $\begingroup$ If I choose $-\sqrt{2}i$, then the choice $\mathbb{C}-[0,1]$ is wrong? If that is so, why? $\endgroup$ – user2820579 May 11 '17 at 18:02
  • $\begingroup$ If you choose $-\sqrt{2}i$, its imaginary part is negative. The question says "positive imaginary part at $z=2$". $\endgroup$ – Robert Israel May 11 '17 at 19:27

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