Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Your formula $(w^2)’=2w$ should be $(w^2)’=2w'w$. So that part of your argument is changed a tiny bit. – Harald Hanche-Olsen Sep 17 '12 at 8:51
up vote 4 down vote accepted

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.

share|cite|improve this answer
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 ? – Chris Sep 18 '12 at 3:38
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. – Harald Hanche-Olsen Sep 18 '12 at 7:40
So to identify the branch, is it enought to show the value a function takes on for a given z? – Chris Sep 19 '12 at 3:17
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.) – Harald Hanche-Olsen Sep 19 '12 at 6:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.