# cv.complex variables – Factoring a complex function such that it is analytic in upper and lower plane

Consider this function $$frac{k^{2}-xi^{2}}{k^{2}+1}$$
which has singularities at $$k=pm i$$, the strips where it is analytic are
$$-1
Let us concentrate on the strip $$-1 The decomposition in this strip yields
$$1-hat{K}(k)=frac{k^{2}-xi^{2}}{k^{2}+1}=frac{A(k)}{B(k)}=frac{k^{2}-xi^{2}}{k-i} frac{1}{k+i}$$
where
$$begin{array}{c} A(k)=frac{k^{2}-xi^{2}}{k-i} \ B(k)=k+i end{array}$$
such that now $$A(k)$$ is analytic for $$k^{prime prime}<0$$ and $$B(k)$$ is analytic for $$k^{prime prime}>-1$$. I worked out it just based on trial and error(Or I would say it is easy to find out such factors in this case).

Now a paper that I am reading claims that for
$$1+widetilde{K}(p)=2 e^{-(c / 2)|p|} cosh left(frac{c}{2} pright)$$
the decomposition
$$1+tilde{K}(p)=frac{K_{+}left(frac{c}{2 pi} pright)}{K_{-}left(frac{c}{2 pi} pright)}$$
in such a way that $$K_{+}left(K_{-}right)$$ has singularities only in the upper (lower) half plane of the complex $$p$$ plane is given by

begin{aligned} K_{+}(q) &=K_{-}^{-1}(-q)=frac{(2 pi)^{1 / 2}}{Gammaleft(frac{1}{2}+i qright)} exp left(-i qleft(1+frac{i pi}{2}-ln (-q+i varepsilon)right)right) end{aligned}

Is there a general process by which they got it?

Also, I have another problem where $$c$$ is now a complex variable instead of real as in this decomposition. How does changing $$c$$ to $$c_r+i c_i$$ change the above factorization?

The necessity for such a factorization appears in study of integral equations that are solvable by Wiener-Hopf method.