Evaluate $\int_C \frac{e^{1/z}}{(z-i)^3}dz$ in the circle $|z|=5$ 
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!
 A: 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.
A: 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 $0$.
A: 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.)

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.
A: 
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.


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!
