MHB Calculating Residues Using Laurent Series

[shen07]

Hi guys,

well i have the problem below,

$$\int_{\gamma(0;1)}\frac{1}{\exp(iz)-1}\mathrm{d}z$$

so it is holormorphic in D'(0,1) as it has a point not holomorphic at z=0.

Taking a Laurent Series in the form $$f(z)=\sum_{n=-\infty}^{\infty}C_n(z-0)^n$$

But i wil get $$\frac{1}{\exp(iz)-1}=\frac{1}{iz+(iz)^2+(iz)^3+\cdots}$$
If i get the coefficient of $$C_-1$$, then i am done. I will have to do a long Division involving $$i$$, Is there any simpler way to do that?? Or please help me the way i am proceeding.

 

[Prove It]

Hi guys,

well i have the problem below,

$$\int_{\gamma(0;1)}\frac{1}{\exp(iz)-1}\mathrm{d}z$$

so it is holormorphic in D'(0,1) as it has a point not holomorphic at z=0.

Taking a Laurent Series in the form $$f(z)=\sum_{n=-\infty}^{\infty}C_n(z-0)^n$$

But i wil get $$\frac{1}{\exp(iz)-1}=\frac{1}{iz+(iz)^2+(iz)^3+\cdots}$$
If i get the coefficient of $$C_-1$$, then i am done. I will have to do a long Division involving $$i$$, Is there any simpler way to do that?? Or please help me the way i am proceeding.

Since the contour is closed and there is only one simple pole, $\displaystyle \begin{align*} z = 0 + 0i \end{align*}$, we can evaluate this using the residue theorem.

$\displaystyle \begin{align*} \int_{\gamma (0, 1)} { \frac{1}{e^{i\,z} - 1}\,dz } &= 2\pi\,i\,\textrm{Res}_{z = 0} \left( \frac{1}{e^{i\,z} - 1} \right) \\ &= 2\pi \, i \lim_{z \to 0} \left[ \left( z - 0 \right) \left( \frac{1}{ e^{i\,z} - 1 } \right) \right] \\ &= 2\pi\,i \lim_{z \to 0} \frac{z}{e^{i\,z} - 1} \\ &= 2\pi \, i \lim_{z \to 0} \frac{1}{ i\,e^{i\,z} } \textrm{ by L'Hospital's Rule} \\ &= 2\pi\,i \left( \frac{1}{i} \right) \\ &= 2\pi \end{align*}$

 

[alyafey22]

Assume the following

$$f(z)=\frac{g(z)}{h(z)}$$

where $h(z)$ has a simple zero at $z_0$ and $g$ is analytic at $z_0$ and $g(z_0)\neq 0 $ then

$$Res(f;z_0) = \lim_{z\to z_0}(z-z_0)\frac{g(z)}{h(z)}=\lim_{z \to z_0}\frac{g(z)}{\frac{h(z)-h(z_0)}{z-z_0}}$$

Now since $h$ is analytic on $z_0$ and $h'(z_0)\neq 0$ because the zero is simple we have

$$Res(f;z_0) =\frac{\lim_{z \to z_0}g(z)}{\lim_{z \to z_0}\frac{h(z)-h(z_0)}{z-z_0}}=\frac{g(z_0)}{h'(z_0)}$$

If on the other hand $g$ has a zero at $z_0$ then the $f$ has a removable singularity at $z_0$. So $Res(f;z_0)=0$.