# Thread: [SOLVED] Find a residue (checking my result and clearing doubts)

1. ## [SOLVED] Find a residue (checking my result and clearing doubts)

Find the residue in $\displaystyle z=0$ of $\displaystyle f(z)=exp\left ( z+ \frac{1}{z} \right )$.

My attempt : I notice that $\displaystyle f$ is analytic except in $\displaystyle 0$. (well I think so! I guess I should show that $\displaystyle f$ is analytic. Is that sufficient that f satisfy C-R equations? Because as you'll see, I need the analyticity of $\displaystyle f$ for the coming part)
I believe that $\displaystyle \int _{\gamma} f(z)dz = 2\pi i= Res f(0)$.
I forgot something very important : why does $\displaystyle \int _{\gamma} f(z)dz = 2\pi i$? I mean, I think there's a theorem that justify it (it seems it's not Cauchy-Goursat theorem) but I forgot which one.

Am I right?

2. Originally Posted by arbolis
Find the residue in $\displaystyle z=0$ of $\displaystyle f(z)=exp\left ( z+ \frac{1}{z} \right )$.

My attempt : I notice that $\displaystyle f$ is analytic except in $\displaystyle 0$. (well I think so! I guess I should show that $\displaystyle f$ is analytic. Is that sufficient that f satisfy C-R equations? Because as you'll see, I need the analyticity of $\displaystyle f$ for the coming part)
I believe that $\displaystyle \int _{\gamma} f(z)dz = 2\pi i= Res f(0)$.
I forgot something very important : why does $\displaystyle \int _{\gamma} f(z)dz = 2\pi i$? I mean, I think there's a theorem that justify it (it seems it's not Cauchy-Goursat theorem) but I forgot which one.

Am I right?
I think (I haven't checked) that z = 0 is an essential singularity of $\displaystyle f(z) = exp\left ( z+ \frac{1}{z} \right )$. To get the residue you'll need to calculate the coefficient of $\displaystyle \frac{1}{z}$ in the Laurent series about z = 0.

3. That's a particular example of what happens when a Taylor series is multiplied by a singular series: the term on $\displaystyle 1/z$ turns out to be an infinite sum sometimes. You have:

$\displaystyle e^{z} e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\sum_{n=0}^{ \infty}\frac{1}{z^n n!}=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{k!}\frac{1}{z^{n-k}(n-k)!}$

But we can figure out which terms have $\displaystyle 1/z$ in them. It's whenever $\displaystyle k-(n-k)=-1$ right? Can you then see:

$\displaystyle \mathop\text{Res}\limits_{z=0} e^{z+1/z}=\sum_{k=0}^{\infty}\frac{1}{k!(k+1)!}$

4. Ah ok so I was all wrong.
Next time I'll think about writing $\displaystyle exp(a(z)+b(z))$ as $\displaystyle exp(a(z)) \cdot exp (b(z))$.
shawsend, I follow you until you stated
Can you then see:

. No, I don't see this.

Ans thanks mr fantastic, I'll try.

5. The powers of z in the double sum have the form $\displaystyle \frac{z^k}{z^{n-k}}=z^{k-(n-k)}$. We want all the 1/z terms or when $\displaystyle k-(n-k)=-1$ or when $\displaystyle n=2k+1$ and when n is that value, the coefficient on 1/z is $\displaystyle \frac{1}{k!(k+1)!}$. I'm just replacing the (n-k) expression in the double sum with (2k+1-k). But since k is going from 0 to n and n is going to infinity, that's the same as k going from 0 to infinity (eventually) and that's why I use the summation expression.

Try and work through the same process for the following:

$\displaystyle \oint_{|z|=1} \cos(z)e^{1/z}dz$

6. Originally Posted by shawsend
The powers of z in the double sum have the form $\displaystyle \frac{z^k}{z^{n-k}}=z^{k-(n-k)}$. We want all the 1/z terms or when $\displaystyle k-(n-k)=-1$ or when $\displaystyle n=2k+1$ and when n is that value, the coefficient on 1/z is $\displaystyle \frac{1}{k!(k+1)!}$. I'm just replacing the (n-k) expression in the double sum with (2k+1-k). But since k is going from 0 to n and n is going to infinity, that's the same as k going from 0 to infinity (eventually) and that's why I use the summation expression.

Try and work through the same process for the following:

$\displaystyle \oint_{|z|=1} \cos(z)e^{1/z}dz$
Thanks for the clarification. I've tried your integral, which gave me $\displaystyle \mathop\text{Res} \cos (z) e ^{\frac{1}{z}}$ in $\displaystyle z=0$ is worth $\displaystyle \sum _{k=0}^{\infty} \frac{(-1)^k}{(2k)!(2k+1)!}$.
I hope I'm right.

7. Very good Arbolis. I think you have it now. Here's the numeric check in Mathematica which expresses the sum in terms of some obscure KevinBel function.

Code:
In[208]:=
Sum[(-1)^k/((2*k)!*(2*k + 1)!),
{k, 0, Infinity}]

NIntegrate[Cos[z]*Exp[1/z]*I*Exp[I*t] /.
z -> Exp[I*t], {t, 0, 2*Pi}]
N[2*Pi*I*((KelvinBei[1, 2] -
KelvinBer[1, 2])/Sqrt[2])]

Out[209]=
(KelvinBei[1, 2] - KelvinBer[1, 2])/
Sqrt[2]

Out[210]=
-4.440892098500626*^-16 +
5.761766462098244*I

Out[211]=
0. + 5.7617664620982385*I

8. Originally Posted by shawsend
That's a particular example of what happens when a Taylor series is multiplied by a singular series: the term on $\displaystyle 1/z$ turns out to be an infinite sum sometimes. You have:

$\displaystyle e^{z} e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\sum_{n=0}^{ \infty}\frac{1}{z^n n!}=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{k!}\frac{1}{z^{n-k}(n-k)!}$

But we can figure out which terms have $\displaystyle 1/z$ in them. It's whenever $\displaystyle k-(n-k)=-1$ right? Can you then see:

$\displaystyle \mathop\text{Res}\limits_{z=0} e^{z+1/z}=\sum_{k=0}^{\infty}\frac{1}{k!(k+1)!}$
Of related interest:

Wolfram|Alpha 1

Wolfram|Alpha 2