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

Laurent's theorem states that a holomorphic function on $R<|z-a|<S$ has a unique expansion ${\sum }_{n\in \mathbb{Z}}{a}_n(z-a{)}^n$, converging absolutely there and uniformly on compact subannuli, withHere $\sin z=z(1-z^2/6+\cdots )$ has a simple zero at zero and no other zero in $|z|<\pi$. Hence $1/\sin z$ has only a simple pole there, giving the stated series with no powers below $z^{-1}$. Multiplying by $\sin z$ and comparing the constant, linear and quadratic coefficients yields

Use the coefficient integrals on the two circles. The annular region between them contains exactly the simple poles $\pi ,-\pi$, and the residue of $1/\sin z$ at each is $1/\cos (\pm \pi )=-1$. Apply the residue theorem to $1/(z^{n+1}\sin z)$ between the circles:For lower indices, set $c_n=0$ because the inner Laurent series has no such terms; the same contour argument then still applies. Since $f$ is odd, its even-index coefficients vanish. A complete explicit list for the requested range isThe subtraction accounts for the two poles crossed; using the inner coefficients unchanged outside their annulus would miss precisely these terms.

For $t>0$, integrate $e^{itz}/(1+z^2)$ over the upper semicircle of radius $R$. The arc tends to zero, since $|e^{itz}|\le 1$ there and the denominator is of order $R^2$. The sole pole is $i$, of residue $e^{-t}/(2i)$, so the real part gives $I(t)=\pi e^{-t}$. The integrand defining $I$ is even in $t$, and $I(0)=\pi$ by the arctangent integral. Hence

For $J$ use $e^{itz}/[z(1+z^2)]$ with a small clockwise semicircular indentation above zero. Its residue at zero is one, so the small arc tends to $-i\pi$. The pole $i$ has residue $-e^{-t}/2$, and the large arc again vanishes. ThusThe imaginary part is the ordinary convergent integral $J(t)$: its apparent singularity at zero is removable, with limit $t$, and its tails are absolutely integrable. Therefore $J(t)=\pi (1-e^{-t})$ for $t>0$. Oddness in $t$ and the value zero at $t=0$ give

Differentiation is valid at every real $t$, including zero. Indeed the mean value theorem bounds the absolute difference quotient of $\sin (tx)/[x(1+x^2)]$ with respect to $t$ by $1/(1+x^2)$, an integrable majorant independent of $t$. Dominated convergence therefore gives $\boxed{J^{\prime }(t)=I(t)\text{ for all }t\in \mathbb{R}}$. In particular both one-sided derivatives at zero equal $\pi$.

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.