Schwarz-Christoffel formula for a half-plane

117 Views Asked by At

I can't understand the example that was given in the book Schwarz-Christoffel Mapping by Tobin Driscoll and Lloyd Trefethen. It's formula 2.5 at page 12. By using Schwarz–Christoffel formula for a half-plane: $$ f(z) = A + C \int^{z} \prod_{k=1}^{n-1} (\zeta-z_{k})^{\alpha_{k}-1} d\zeta $$ he concludes that for $n=1$, with $w_{1}=\infty$ and $\alpha_{1}=-1$, we have a line (that I can understand), and gets to:

$$ f(z) = A + Cz $$ I know that he can absorb constants into $C$, nevertheless I can't understand why we have anything in this integrand because the product is empty for $n=1$, therefore it should be zero, and we would have $f(z) = A$.

1

There are 1 best solutions below

0
On BEST ANSWER

An empty sum is zero, because it's the quantity which has no effect when you add it. Likewise, an empty product is naturally defined to be $1$, not $0$. Thus the integrand is one and $f(z) = A + Cz$ rather than $A$.