Let $f(x) = (z-\frac{1}{z})$ , $z \in \mathbb{C}$\ {${0}$}

Let $F$ be a holomorphic branch of the square root of $f$ that is defined at $z=2$ and has the value $\sqrt{3/2}$ there.

GIve an explicit expression for $F$ as a composition of standard functions ( exponential, logarithm etc)

My attempt : $$F(z)^2 = (z-\frac{1}{z})$$ $$F(z) = (z-\frac{1}{z})^{\frac{1}{2}}$$ $$= exp(\frac{!}{2} (log (z-\frac{1}{z})))$$

I am stuck here. I know $f$ is real valued when $z$ is real valued.


1 Answer 1


Presumably, the question asks you to show that the branch of the logarithm you've defined is holomorphic around $2$. Take a ball of appropriate radius around $2$,and argue that if a function $f$ is non - zero in a simply connected domain, then it has a well-defined holomorphic branch of a logarithm there.

$\textbf{Edit With More Details}:$ First, recall that if $U$ is a simply-connected domain and $f:U\to \mathbb{C}\setminus\{0\}$ is analytic, then there exists a well-defined analytic branch of the logarithm of $f$ on $U$. Secondly, for the problem at hand, observe that the given $f$ equals $0$ at $z=\pm1$, therefore $f$ is non-zero in a ball of radius $\dfrac{1}{2}$ around $2$. Now combine these two facts.

  • $\begingroup$ Can you help me with details? I am not clear on your solution. Thanks $\endgroup$
    – Rusty
    Jul 5, 2016 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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