1
$\begingroup$

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?

  • 0
    Your formula $(w^2)’=2w$ should be $(w^2)’=2w'w$. So that part of your argument is changed a tiny bit.2012-09-17

1 Answers 1

4

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.

  • 0
    Yes, exactly, at least in this case. The theory of [analytic continuation](http://en.wikipedia.org/wiki/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.)2012-09-19