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

What I’ve gotten so far:

I note that each branch of $w=(z(1-z))^{1/2}$ satisfies $w^2=(z(1-z))$ and that $f(z)$ is continuous on $\mathbb{C} \setminus [0,1]$. Since $(w^2)’=2w$ is not zero for $w\neq 0$, the continuous inverse branch $(z(1-z))^{1/2}$ is analytic. Differentiating $w= (z(1-z))^{1/2}$ we obtain $dw/dz=(1-2z)/2(z(1-z))^{1/2}$ Now I know that $dz/dw=1/(dw/dz)$ so I know $f’(z)$ but I’m a little confused about specifying the right branch, any clues?

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  • $\begingroup$ Your formula $(w^2)’=2w$ should be $(w^2)’=2w'w$. So that part of your argument is changed a tiny bit. $\endgroup$ – Harald Hanche-Olsen Sep 17 '12 at 8:51
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It is better to work with $w^2=z(1-z)$ and to differentiate that, with the result $2ww'=1-2z$. Write that as $$f'(z)=\frac{1-2z}{2f(z)},$$ and it should now be obvious how the choice of the branch of the square root in the expression of $f(z)$ affects the choice for $f'(z)$. Consider $z=2$ and compute everything in sight. I trust you can take it from there.

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  • $\begingroup$ So since f'(z)=-3/2(2)^(1/2) and f(2) has positive imaginary part, f'(z) is the branch that maps to Rez>0 ? $\endgroup$ – Chris Sep 18 '12 at 3:38
  • $\begingroup$ I find that a bit confused. $f(2)$ is the square root with positive imaginary part of $-2$, so $f(2)=i\sqrt2$. Plug that into the formula to get $f'(2)=3i/(4\sqrt2)$, which is enough to identify the branch. $\endgroup$ – Harald Hanche-Olsen Sep 18 '12 at 7:40
  • $\begingroup$ So to identify the branch, is it enought to show the value a function takes on for a given z? $\endgroup$ – Chris Sep 19 '12 at 3:17
  • $\begingroup$ Yes, exactly, at least in this case. The theory of analytic continuation says that it is enough to specify the value in a neighbourhood of a given point in general. In this case, as there are only two branches and they take different values at $z=2$, specifying $f'(2)$ is sufficient. But note that $f'(1/2)=0$ for both branches, so $z=1/2$ is not a good point to use for this purpose. (You don't need the full power of the theory of analytic continuation for this, just uniqueness of continuation along a path.) $\endgroup$ – Harald Hanche-Olsen Sep 19 '12 at 6:56

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