Where is the mistake $\oint_C \frac{e^\frac{1}{z}}{z^2-1}dz$

Question

Evaluate $\oint_C \frac{e^\frac{1}{z}}{z^2-1} dz$ where $C$ is the locus of $z$ satisfying $|z-1|<3/2$ (answer: $iπ/e$).

My attempt at an answer comes nowhere close to this given result:

The two residues inside our contour are at $z=0$ and $z=1$. By Cauchy's residue theorem, the value of the integral is $2\pi i$ times the sum of these two residues.

At $z=ε, ε\ll 1$, we have: $$\frac{1}{ε^2-1}(1+1/ε+...) \rightarrow \frac{1}{-1}(1+1/ε+...) \rightarrow \text{Res} = -1$$

At $z=1+ε$, we have: $$ e^\frac{1}{1+ε} \frac{1}{(ε)(2+ε)} \rightarrow \frac{1}{ε} e/2 \rightarrow \text{Res} = e/2$$

So my answer is $2πi (\frac{e}{2}-1)$ - apparently this is completely wrong though! The only way I can see to get the given answer is if we include the resiude at $z=-1$. Have I completely misunderstood something?

Answer 1

The mistake is from here

At $z=ε, ε\ll 1$, we have: $$\frac{1}{ε^2-1}(1+1/ε+...) \rightarrow \frac{1}{-1}(1+1/ε+...) \rightarrow \text{Res} = -1$$

At $z=0$, do the Laurent series:

$$\frac1{z^2-1}e^{1/z}=-\sum_{k=0}^\infty \frac1{k!}z^{-k}\sum_{n=0}^\infty z^{2n} $$

take the coefficient for the $z^{-1}$ term,

$$\text{Res}(z=0)=-\sum_{k=0}^\infty \frac1{(2k+1)!}=-\sinh1=\frac{1}{2e}-\frac e2$$

So we get

$$\text{Res}(z=0)+\text{Res}(z=1)=\frac{1}{2e}-\frac e2+\frac e2=\frac{1}{2e}$$

which matches the answer.

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Best Method is to consider the area enclosed by the contour outside $|z-1|\ge \frac32$, which includes a simple pole at $z=-1$ and the singuarity $z=\infty$. Note that $z=\infty$ is analytic in your case. So the residue at $z=\infty$ is zero. And you immediately get the answer by only computing the residue at $z=-1$. But be careful to add an additional $-$ sign in this case because the contour is clockwise with respect to the area outside $|z-1|\ge \frac32$.

Answer 2

The error is in the computation of the residue of the integrand at $z = 0$, i.e., in this computation:

At $z=ε, ε\ll 1$, we have: $$\frac{1}{ε^2-1}(1+1/ε+...) \rightarrow \frac{1}{-1}(1+1/ε+...) \rightarrow \text{Res} = -1$$

Denote $$f(z) := \frac{\exp \frac1z}{z^2 - 1} .$$

Hint When $|z| < 1$, $$\frac{1}{z^2 - 1} = -(1 + z^2 + z^4 + \cdots),$$ so the Laurent expansion of $\frac{\exp \frac1z}{z^2 - 1}$ in $\{0 < |z| < 1\}$ is given by expanding the product $$-(1 + z^2 + z^4 + \cdots) \left(1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3!z^3} + \cdots\right) .$$ In particular, the coefficient of $\frac{1}{z}$ is $$\operatorname{Res}\left(f, 0\right) = -\left(1 + \frac{1}{3!} + \frac{1}{5!} + \cdots\right) .$$

Additional hint We recognize the quantity in parentheses as the Taylor expansion of $\sinh z = \frac{1}{2}(\exp z - \exp(-z))$ evaluated at $z = 1$, so the residue at $0$ is $$\operatorname{Res}\left(f, 0\right) = -\sinh 1 = \frac{1}{2}\left(\frac{1}{e}-e\right) .$$

Answer 3

Alternatively to the other answers, another way would be the change of variables $\zeta = \frac{1}{z}$, i.e. $d z = -\frac{1}{\zeta^2}d\zeta$.

The only thing one has to be careful with is the contour one integrates over: Denote by $\Gamma$ the circle described by $|z-1|=\frac{3}{2}$, then the conformal change of coordinates above maps $\Gamma$ again to a circle $\Gamma'$. As one easily verifies, $\Gamma$ encircled $z=1$ once in anti-clockwise direction and did not encircle $z=-1$, while in the image coordinates, $\Gamma'$ encircles $\zeta = -1$ in the clockwise direction, but not $\zeta=1$.

Thus, one has:

\begin{align}\int\limits_\Gamma \frac{e^{\frac{1}{z}}}{z^2-1}dz &= -\int\limits_{\Gamma'} \frac{e^\zeta}{\frac{1}{\zeta^2}-1}\frac{1}{\zeta^2}d \zeta\\ & = \int\limits_{\Gamma'} \frac{e^\zeta}{\zeta^2-1}d\zeta\\ & = w(\Gamma',-1)\cdot 2\pi i \cdot \operatorname{Res}\left(\frac{e^\zeta}{\zeta^2-1}, -1\right)\\ & = -2\pi i\cdot \frac{e^{-1}}{-2}\\ & = \frac{\pi i}{e} \end{align} In the third line, $w(\Gamma',-1)=-1$ denotes the winding number of $\Gamma'$ around $-1$ (it is negative because $\Gamma'$ is going in the clockwise direction).