Answer to Question #232174 in Complex Analysis for Fozia Sayda

Question

Find the residue of the function f(z)=1/z(e^z-1)

Answer 1

 If $f(z)$ has a pole of order $k$ at $z=z_0$ then

$$rez(f,z_0)=\lim\limits_{z\to z_0}\big(\dfrac{1}{(k-1)!}\dfrac{d^{k-1}}{dz^{k-1}}((z-z_0)^kf(z))\big)$$

$$e^z=1+\dfrac{z}{1!}+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+...$$

$$e^z-1=\dfrac{z}{1!}+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+...$$

$$\lim\limits_{z\to0}(zf(z))=\lim\limits_{z\to0}(z(\dfrac{1}{z(e^z-1)}))=\lim\limits_{z\to0}(\dfrac{1}{e^z-1})$$

$$=does\ not \ exist$$

$$\dfrac{1}{(2-1)!}\dfrac{d^{2-1}}{dz^{2-1}}((z-0)^2f(z))$$

$$=\dfrac{d}{dz}(z^2(\dfrac{1}{z(e^z-1)}))$$

$$=\dfrac{d}{dz}(\dfrac{z}{e^z-1})$$

$$=\dfrac{e^z-1-ze^z}{(e^z-1)^2}$$

$$rez(f,0)=\lim\limits_{z\to0}\big(\dfrac{e^z-1-ze^z}{(e^z-1)^2}\big)$$

$$=\lim\limits_{z\to0}\big(\dfrac{e^z-e^z-ze^z}{2e^z(e^z-1)}\big)$$

$$=\lim\limits_{z\to0}\big(\dfrac{-z}{2(e^z-1)}\big)$$

$$=\lim\limits_{z\to0}\big(\dfrac{-1}{2e^z}\big)=-\dfrac{1}{2}$$

$$rez(f,0)=-\dfrac{1}{2}$$