Define a branch of $(z^2 − 1)^{1/2}$ which is analytic in the unit disk.

Question

Hint: $z^2 − 1 = (z + 1)(z − 1)$.

I'm really struggling with this question. I understand that for this function to be analytic it has to be differentiable in some neighbourhood, but I have no idea where to start or even how to find/pick a branch.

Answer 1

First solution. We use the following lemma

Lemma. Let $\Omega\subset\mathbb C$ be an open region. If $a,b$ lie in the same connected component of $\,\mathbb C\cup\{\infty\}\setminus\Omega$, then $$ f(z)=\frac{z-a}{z-b} $$ possesses an analytic logarithm in $\Omega$.

Using this lemma, let $g\in\mathcal H(\mathbb D)$, where $\mathbb D$ is the unit disk, the logarithm of $\,f(z)=\dfrac{z-1}{z+1}$, and set $$ h(z)=(z+1)\exp\big(\tfrac{1}{2}g(z)\big). $$ Then $$ \big(h(z)\big)^2=(z+1)^2\exp\big(g(z)\big)=(z+1)^2\cdot\dfrac{z-1}{z+1}=z^2-1. $$ Proof of Lemma. Observe that $$ \frac{f'(z)}{f(z)}=\frac{1}{z-a}-\frac{1}{z-b}. $$ Hence, if $\gamma$ is a closed curve in $\Omega$, then $$ \int_\gamma \frac{f'}{f}=\mathrm{Ind}_\gamma(a)-\mathrm{Ind}_\gamma(b)=0, $$ since $a,b$ lie in the same connected component of $\,\mathbb C\cup\{\infty\}\setminus\Omega$.

Thus $\frac{f'}{f}$ possesses an holomorphic anti-derivative in $\Omega$, the analytic logarithm of $f$.

Second solution. Set $f(z)=g(z)h(z)$, where $g(z)=\sqrt{z-1}$, defined in $\mathbb C\smallsetminus [1,\infty)$ and $h(z)=\sqrt{z+1}$, defined in $\mathbb C\smallsetminus (-\infty,-1]$.

Now, how do we define $h(z)=\sqrt{z+1}$ in $\mathbb C\smallsetminus (-\infty,-1]$?

Every $z\in \mathbb C\smallsetminus (-\infty,-1]$ can be written UNIQUELY as $z=-1+r\mathrm{e}^{i\vartheta}$, with $r>0$ and $\vartheta\in (-\pi,\pi)$. then show that the function $$ H(r,\vartheta)=r^{1/2}\mathrm{e}^{i\vartheta/2}, $$ for $r>0$ and $\vartheta\in (-\pi,\pi)$. You can do it expressing Cauchy-Riemann equations with respect to $r$ and $\vartheta$ (See $(3)$ in this paper.)

Answer 2

Define $$f(z)= \sqrt{z^2-1}$$

Then function has two branch points at $z,=1,z=-1$, So it can be extended to the whole complex plane except at two branch cuts chosen properly.

you can choose the branch cut at $z^2-1\geq 0$ os $z\geq 1 \text{ and } z\leq -1$.

This is like defining the argument as a singly valued function by choosing

$$\text{arg}(z) \in (0,2\pi]$$

Hence we here are choosing the half line $z\geq 0$ to be the branch cut . Simiarily for the function

$$f(z)=z^{\frac{1}{2}}=e^{\frac{1}{2} \text{Log}(z)}$$

Here we are choosing the principle logarithm so the function is the principle root and is analytic for $z \in \mathbb{C}/(-\infty,0]$.

Answer 3

If we make the branch cut along $\{x\in\mathbb{R}:|x|\ge1\}$ and define $$ f(z)=\frac{\pi i}{2}+\frac12\int_0^z\left(\frac1{w-1}+\frac1{w+1}\right)\,\mathrm{d}w $$ then $f$ is well-defined since we can't circle either singularity with the branch cut. Furthermore, $$ \sqrt{z^2-1}=e^{f(z)} $$

Answer 4

it s very simple if you use the graphics, so you use $z-1 =r_1e^{i\theta_1}$, and so $z+1 =r_2e^{i\theta_2}$, than the function become $f (z)=\sqrt{r_1r_2}e^{i \frac{\theta_1+\theta_2}{2}}$, study the argument by using a graphics.