Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume that we have two meromorphic functions $f(z,w)$ and $g(z,w)$, where $z$ and $w$ are complex (we are interested only in behavior on compact sets). Fix $z$ and assume that the sets of poles of $f(z,\cdot)$ and $g(z,\cdot)$ do not intersect. Then we can choose a contour $\gamma$ such that all the poles of $f(z,\cdot)$, but none of the poles of $g(z,\cdot)$, lie inside this contour, and calculate the integral $I(z)=\int_\gamma f(z,w)g(z,w)dw$.

Now assume that the poles of $f(z,\cdot)$ and $g(z,\cdot)$ coincide for certain $z$. Then (at least under certain assumptions) the function $I(z)$ is meromorphic, say, by Weierstrass Preparation. The question is: does this construction come up anywhere in the classical complex analysis, and if so, where can one read about it?

Update 1: One case where this construction comes up is when you try to construct the inverse $$ (A(z)\otimes 1+1\otimes B(z))^{-1} $$ from the inverses $f(z,w)=(A(z)-w)^{-1}$ and $g(z,w)=(B(z)+w)^{-1}$. Here $A(z)$ and $B(z)$ are holomorphic families of linear operators on two finite-dimensional spaces $V$ and $W$ and $A\otimes 1+1\otimes B$ acts on $V\otimes W$.

Update 2: what I could come up with so far. Assume that at $z=0$, both $f(z,\cdot)$ and $g(z,\cdot)$ have exactly one same pole at $w=0$. Let $P(z,w)$ and $Q(z,w)$ be the polynomials obtained from $1/f$ and $1/g$ near $(0,0)$ via Weierstrass Preparation Theorem. (They are polynomials in $w$ with coefficients holomorphic in $z$.). Using Euclid's algorithm, we find two polynomials $p(w)$ and $q(w)$ such that $Pp+Qq$ is identically zero in $w$ at $z=0$. With luck, the coefficients of $Pp+Qq$ are not identically zero in $z$; then we can use that $(Pp+Qq)/(PQ)$ is easy to integrate over the contour described above to get a meromorphic expansion for $I(z)$ at $z=0$.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.