Evaluate $\int_C \frac{e^{1/z}}{(z-i)^3}dz$ in the circle $|z|=5$

Question

Evaluate $\int_C \frac{e^{1/z}}{(z-i)^3}dz$ in the circle $|z|=5$

I want to check that if my solution is correct.

Since $1/z$ is not analytic at $z=0$ and $1/(z-i)^3$ is not analytic at $z=i$, hence the function $f(z) =\frac{e^{1/z}}{(z-i)^3}$ has singularities at $z=0$ and $z=3$.

Notice that both of them are inside the circle $|z|=5$.

So we can use residue at infinity to evaluate this integral.

We have:

$f(1/z)= \frac{e^z}{(1/z-i)^3} = \frac{z^3e^z}{(1-iz)^3} \implies \frac{1}{z^2}f(1/z) = \frac{ze^z}{(1-iz)^3}.$

We must find the residue of $\frac{1}{z^2}f(1/z)$ at $z=0$. But notice that this function is analytic at this point. Hence its Laurent expansion is equal to its Taylor expansion. Its principal part is then zero, so we have

$\mbox{Res}_{z = 0}\frac{1}{z^2}f(1/z) = 0$.

So, finally: $\int_C \frac{e^{1/z}}{(z-i)^3}dz = 2\pi i*0 = 0.$

Is this solution correct? I am asking because I did not need to find any series expansion!

Answer 1

I thought it might be instructive to present an approach that is equivalent to using the residue at infinity, but perhaps a bit simpler in this case. To that end we proceed.

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Inasmuch as the singularities of the integrand lie inside the unit circle, Cauchy's Integral Theorem guarantees that for $R>5$

$$\oint_{|z|=5}\frac{e^{1/z}}{(z-i)^3}\,dz=\oint_{|z|=R}\frac{e^{1/z}}{(z-i)^3}\,dz$$

Now, we have the simple estimate

$$\begin{align} \left|\oint_{|z|=R}\frac{e^{1/z}}{(z-i)^3}\,dz\right|&=\left|\int_0^{2\pi} \frac{e^{\frac1Re^{-i\phi}}}{(Re^{i\phi}-i)^3}\,iRe^{i\phi}\,d\phi\right|\\\\ &\le \frac{2\pi R^2}{(R-1)^4} \end{align}$$

Finally, letting $R\to \infty$, we find

$$\oint_{|z|=5}\frac{e^{1/z}}{(z-i)^3}\,dz=0$$

And we are done!

Answer 2

Yes. Or use the following more general argument. Whenever the integrand is regular outside the contour, we can make the contour progressively larger (radius $R$) without altering the integral, by Cauchy's thm, and if the integrand is $o(1/R)$ on the contour we will always get zero, as by the Bounding Lemma the integral is less than circumference of contour $\times$ maximum of integrand, which goes as $2\pi R \times o(1/R) \to 0$. In effect this amounts to changing variable $z\mapsto 1/z$ and using analyticity.

Answer 3

OK, i will type an answer, after the comment. The result $0$ is correct. We change variables, $w=1/z$. The contour $C$, circle with radius $5$ goes into the circle $C'$ with radius $1/5$. Then we have $$ \begin{aligned} \int_C\frac {e^{1/z}}{(z-i)^3}\; dz &= \int_{C'}\frac{e^w}{\left(\frac 1w-i\right)^3}\; \left(-\frac 1{w^2}\right)\;dw \\ &= -\int_{C'}\frac {w^3e^w}{w^2(1-iw)^3}\;dw \\ &= -\int_{C'}\frac {we^w}{(1-iw)^3}\;dw \ . \end{aligned} $$ There is no pole inside $C'$. (The pole $w=-i$ is outside.) So the integral is zero, Residue Theorem.

Numerically, pari/gp:

? intnum( t=0, 2*Pi, 
          exp( (cos(t)-I*sin(t))/5 ) 
               / (5*(cos(t)+I*sin(t)) - I)^3 
               * 5*(-sin(t)+I*cos(t)) )
%1 = 1.4190890450537301016459116208947201222 E-21 
   - 1.9571617134379963221261930053722380946 E-21*I

(Manually broken lines to fit in the window.)

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This is of course better than computing the two residues of the initial function (in $0$ and in $i$), which are in $i$...

sage: var('z');
sage: f = exp(1/z) / (z-i)^3
sage: f.residue(z==i)
(I + 1/2)*e^(-I)

computed with sage, and in $0$ with the hand $$ \begin{aligned} &\text{Coeff}_{1/z} \frac {e^{1/z}}{(z-i)^3} \\ &= \text{Coeff}_{1/z} -i\cdot e^{1/z}\cdot\frac 1{(1+iz)^3} \\ &=\text{Coeff}_{1/z} -i\cdot \left( \frac 1{0!} +\frac 1{1!}\frac 1z +\frac 1{2!}\frac 1{z^2} +\frac 1{3!}\frac 1{z^3} +\dots\right) \\ &\qquad\cdot \left( \binom 22 +\binom 32 (-iz) +\binom 42(-iz)^2 %+\binom 52(-iz)^3 +\dots \right) \\ &= -i\left( \frac 1{1!}\binom 22 +\frac 1{2!}\binom 32 (-i) +\frac 1{3!}\binom 42 (-i)^2 %+\frac 1{4!}\binom 52 (-i)^3 +\dots \right) \\ &=\dots \end{aligned} $$ and in the above sum each $\frac 1{(n+1)!}\binom{n+2}2 =\frac 1{(n+1)!}\cdot\frac{(n+2)!}{n!2!}=\frac{(n+2)}{n!2!}$ has to be explained. We split the nummerator in two parts, one with $n$, one with $2$, and will get some parts of the exponential series in $-i$. One of them needs the aid of the missing $1/0!$, so it is acceptable that we get the killing contribution for the first residue.

Answer 4

We have, $\displaystyle f\left(\frac 1z\right)=\frac{z^3e^z}{(1-iz)^3}$.

Now, $\displaystyle Res(f(z),\infty)=Res\left(\frac{1}{z^2}f(1/z),0\right)=Res\left(\frac{ze^z}{(1-iz)^3},0\right)=0 \text{ as it is analytic at $z=0$.}$

Now, Sum of the residues at the finite poles and the residue at infinity is $0$.

So, $Res(f,0)+Res(f,i)=-Res(f,\infty)=0.$

Therefore by Cauchy's residue Theorem integral value is $0$.