# To show that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$ is analytic in the plane minus [-1,1]

By the definition of logarithm branch, we set that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$ However to show that the $\sqrt{z^2-1}$ is analytic in the entire plane minus the interval [-1,1], it is better to prove firstly $\sqrt{z^2-1}$ is continuous along the interval (-∞,1) and then show the above statement by applying Morera's theorem. (My course advisor said to me)

However, when I follow the instruction of my advisor, I just get stuck for solving it. He said that when Im z>0 the path of integration is in the upper half-plane, and vice versa. I understood that statement. Therefore, to prove that limit value of f(z) for z -> x for $x \in (-∞, -1)$, we must establish that the same limit exists also when we approach x to upper half-plane or the lower half-plane.

But I don't understand that the difference between these limits equals to the $\int_C \frac{2\zeta}{\zeta^2-1}$ (C is regular closed curve surrounding $\zeta=+1,-1$ . Why does that differnce is followed by above integral form, and why does the contour is defined as above? I barely grasp the concept..........

Also, assuming that I understand that thing, there is still problematic to complete the proof. By the argument principle, since the $z^2-1$ has its zero in C by +1 and -1, $\int_C \frac{2\zeta}{\zeta^2-1}d\zeta=2 \pi i \times 2 = 4 \pi i$ , Then, how we conclude that limit value exists? My professor just said me that $e^{w/2}=e^{(w+4\pi i)/2}$ but I cannot understand why that thing satisfies. After that, how to applying Morera to complete this things?

Your suggestion will help me !!!!! thanks

• May 15 '13 at 5:20

This is similar to Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$
1. Introduce $f(z)=\frac12\log (z^2-1)$ as a holomorphic function in the domain $\mathbb C\setminus (-\infty,1]$.
2. Show that $\exp(f)$ extends continuously to the interval $(-\infty,-1)$.
3. Use Morera's theorem to show that $\exp(f)$ is holomorphic in $\mathbb C\setminus [-1,1]$.
Step 1 uses the fact that $\mathbb C\setminus (-\infty,1]$ is simply-connected and does not contain the points $\pm 1$ where $f$ has singularities.
Step 2, on which you got stuck, is based on the fundamental theorem of calculus: $f(b)-f(a)=\int_\gamma f'(\zeta)\,d\zeta$ where $\gamma$ can be any smooth curve that goes from $a$ to $b$. Notice that $f'(\zeta)= \zeta/(\zeta^2-1)$. Integrating $f'(\zeta)$ along a simple closed curve that passes through $x_0\in (-\infty,-1)$, we find that the integral is $2\pi i$. This means that $f$ takes on different values when approaching $x_0$ from above and from below. But since the exponential function is $2\pi i$ - periodic, it follows that $\exp(f)$ has the same limit when approaching $x_0$ from above and from below. This is what we wanted.