# Doubts in the solution of a Riemann Hilbert problem

Consider the following Riemann-Hilbert problem as given on Page 9 of this paper: $$\Phi^+(t)- \Phi^-(t) = 2u(t)$$ $$\Phi^+(t)+ \Phi^-(t) = \frac{P}{\pi i } \int_{t_1}^{t_2} \frac{d\zeta}{\zeta} u(\zeta) \frac{\zeta +t}{\zeta - t} = E(t - t^{-1})$$ where $$E\in \mathbb{R}$$, $$P$$ denotes the principal value of the integral, and $$\Phi^{\pm}(t)$$ denote the values of the function $$\Phi$$ as the point $$t$$ on the counterclockwise arc $$(t_1, t_2)$$ is approached from the left and the right respectively. The function $$u(t)$$ takes the boundary values $$u(t_1) = u(t_2) = 0$$. We have an additional constraint here as well:

$$\Phi(\infty) = -1.$$

1. The author then proceeds to write down the solution of this problem which I am confused by. The first confusion is the claim is that "the function $$h(z) = \left[ (z-t_1)(z-t_2)\right]^{1/2}$$ solves the problem $$h^+ + h^- =0$$ upto an entire function." I am not able to see the same.

The author then writes the ansatz for the function $$\Phi(z) = h(z)H(z)$$. Using the Plemelj formula and calculating the residues, the author then obtains the expression for $$H(z)$$ as:

$$H(z) = \frac{E}{2} \left[ \frac{z-z^{-1}}{\sqrt{(z-t_1)(z-t_2)}}+1 + \frac{1}{z}\right]$$

1. Is there an argument why the minus sign in the first term's numerator of the above equation gets converted into a plus sign in the following equation for $$\Phi(z)$$?

$$\Phi(z) = \frac{E}{2} \left( z+ \frac{1}{z}\right) + \frac{E}{2} \left(\frac{1}{z} + 1\right) \sqrt{(z-t_1)(z-t_2)}$$

1. Using the above equation the author substitutes $$\Phi(\infty) = -1$$ to obtain:

$$\frac{1-\cos \alpha}{2} = \frac{1}{E}$$ This follows if I set $$z = e^{i\alpha}$$. However it does not seem reasonable to do so when $$z=\infty$$. Is there some other way to see this?