Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 4E (Further Analysis)

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.

The slit plane is connected and contains $D$. If two holomorphic functions on it extend the same $f$, their difference vanishes on the nonempty open disc $D$. The identity theorem forces their difference to vanish on the entire slit plane. Thus there is at most one analytic extension of the chosen $f$. Choosing the opposite square root initially changes that chosen function; it does not produce two extensions of the same function.

Suppose an analytic extension $F$ existed on the punctured plane. The identity theorem would extend $F^2=z$ from $D$ to that connected domain, so $F$ would never vanish and differentiation would give $2F^{\prime }/F=1/z$. Integrate around the positively oriented unit circle. The argument principle says $\int F^{\prime }/F$ is $2\pi i$ times the integer winding number of the closed curve $F(e^{it})$ about zero, whereasThis would make that winding number $1/2$, a contradiction. Hence no such analytic extension exists. Equivalently, continuation of a square root once around zero reverses its sign, preventing a single-valued holomorphic function on the punctured plane.