Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Analysis / 6

The complex fundamental theorem of calculus says that for a holomorphic antiderivative $F$ and a piecewise $C^1$ path $\gamma$ from $a$ to $b$, one has ${\int }_{\gamma }{F}^{\prime }(z)dz=F(b)-F(a)$. The integral is therefore zero on a closed path; the antiderivative need only be defined on a neighborhood of that path, rather than on all its interior.

The assumptions make $F(z)=z/f(z{)}^n$ holomorphic near $\gamma$. Differentiate:Integrate this derivative around the closed path to obtain

For a positively oriented simple closed piecewise smooth curve, a point $a$ inside it and a function $h$ holomorphic on a neighborhood of the curve and its interior, Cauchy's differentiated integral formula isThe printed denominator uses $n+1$ while the derivative and factorial use $m$. The standard formula requires these indices to agree; the intended exponent in this display is $m+1$. The private transcription preserves the mismatch exactly.

For the following task, write $f(z)=zq(z)$ near zero with $q(0)=f^{\prime }(0)\ne 0$. Thus $H(z)=(z/f(z){)}^n=q(z{)}^{-n}$ extends holomorphically at zero, and is holomorphic inside the specified contour because $f$ has no other zeros there. Now $f(z{)}^{-n}=H(z)/z^n$. Apply the correctly indexed formula with derivative order $n-1$ to getPositive winding one supplies the orientation and multiplicity used in this expression.

Choose a small positively oriented circle $|w|=\rho$ in $U$ within the convergence disk of the inverse series. Cauchy's coefficient formula givesLet $\gamma =g(\{|w|=\rho \})$. The biholomorphism property makes it a simple positively oriented closed curve about zero. Under $w=f(z)$, $g(w)=z$ and $dw=f^{\prime }(z)dz$, soThe last equality is (a). Apply (b), givingThis is the local Lagrange inversion theorem; the simple zero ensures that the expression being differentiated has a removable singularity and a nonzero value at zero.