Singularities and Residues

Multiply the series eᶻ $=\sum z^m/m$! and $e^{1/z}$ $=\sum$ $z^{-k}/k$!; the coefficient of $z^{-1}$ comes from $k=m+1:{\sum }_{m}1/(m!(m+1)!)$.

Casorati–Weierstrass asserts density; surjectivity onto $\mathbb{C}$ minus at most one point is the far stronger Great Picard theorem, and $e^{1/z}$ (never 0) shows 'all of ${\mathbb{C}}^{\prime }$ can fail. Boundedness would make the singularity removable by Riemann's theorem.

(1) Near 0 the factor 1/(z − 1) is analytic and non-zero, so the $z^2$ in the denominator gives a pole of order exactly 2. (2) Fails — z = 1 is a simple pole, since $1/z^2$ is analytic and non-zero there. (3) Res(f, 1) = $li{m}_{z\to 1}$ $(z-1)f(z)=1/1^2=1.(4)$ For the residue at 0 expand $1/(z-1)=-(1+z+z^2+\cdots )$, so $f(z)=-1/z^2-1/z-1-\cdots$ and Res(f, 0) = −1, not 1. As a check, the residues sum to 0, consistent with f decaying like $1/z^3$ at infinity.

(1) sin z/z → 1, and a bounded isolated singularity is removable (Riemann). (2) The Laurent series $\sum$ $z^{-n}/n$! has infinitely many negative-power terms. $(3)1-\cos z=z^2/2-z^4/24+\cdots =z^2\times ($analytic and nonzero at 0), so the reciprocal has a double pole. (4) fails: z·$e^{1/z}$ = z + 1 + 1/(2!z) + ⋯ still has infinitely many negative powers — multiplying an essential singularity by zᵏ never produces a pole.

$I_k=2\pi i\cdot ($coefficient of $z^{k-1}$ in eᶻ$)=2\pi i/(k-1)$! for k ≥ 1 and 0 for k ≤ 0. These values decrease to 0, so (3) and (4) fail.

Residues at ±i are 1/(2i) and −1/(2i); they sum to 0.

The singularities are 0 (simple) and 2 (double); only 0 lies inside |z| = 1. Res$(f,0)=(0+1)/(0-2{)}^2=1/4$, so the integral is $2\pi i\cdot (1/4)=\pi i/2$. At the double pole: Res(f, 2) = d/dz$[(z^2+1)/z]$ at $z=2=(1-1/z^2)$|${}_2=3/4$. All four statements hold — Part C does occasionally do that.

Both simple poles ±i lie inside |z| = 2, with residues 1/(2i) and −1/(2i) (from 1/(2z) at each pole). They cancel, so the integral is 0 — consistent with the integrand decaying like $1/z^2$ at infinity.