Jtdylktuy

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

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

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 $2pi R times o(1/R) to 0$. In effect this amounts to changing variable $zmapsto 1/z$ and using analyticity.

We have, $displaystyle fleft(frac 1zright)=frac{z^3e^z}{(1-iz)^3}$.

Now, $displaystyle Res(f(z),infty)=Resleft(frac{1}{z^2}f(1/z),0right)=Resleft(frac{ze^z}{(1-iz)^3},0right)=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 $0$.

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
$$
begin{aligned}
int_Cfrac {e^{1/z}}{(z-i)^3}; dz
&=
int_{C'}frac{e^w}{left(frac 1w-iright)^3}; left(-frac 1{w^2}right);dw
\
&=
-int_{C'}frac {w^3e^w}{w^2(1-iw)^3};dw
\
&=
-int_{C'}frac {we^w}{(1-iw)^3};dw
.
end{aligned}
$$
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
$$
begin{aligned}
&text{Coeff}_{1/z}
frac {e^{1/z}}{(z-i)^3}
\
&=
text{Coeff}_{1/z}
-icdot e^{1/z}cdotfrac 1{(1+iz)^3}
\
&=text{Coeff}_{1/z}
-icdot
left(
frac 1{0!}
+frac 1{1!}frac 1z
+frac 1{2!}frac 1{z^2}
+frac 1{3!}frac 1{z^3}
+dotsright)
\
&qquadcdot
left(
binom 22
+binom 32 (-iz)
+binom 42(-iz)^2
%+binom 52(-iz)^3
+dots
right)
\
&=
-ileft(
frac 1{1!}binom 22
+frac 1{2!}binom 32 (-i)
+frac 1{3!}binom 42 (-i)^2
%+frac 1{4!}binom 52 (-i)^3
+dots
right)
\
&=dots
end{aligned}
$$
and in the above sum each $frac 1{(n+1)!}binom{n+2}2
=frac 1{(n+1)!}cdotfrac{(n+2)!}{n!2!}=frac{(n+2)}{n!2!}$ 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.

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

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

Finally, letting $Rto infty$, we find

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

And we are done!