Using Cauchy Integral Formula for Laurent Series Coefficients

[cbarker1]

Dear Everyone,

I am wondering how to use the integral formula for a holomorphic function at all points except a point that does not exist in function's analyticity. For instance, Let $f$ be defined as $$f(z)=\frac{z}{e^z-i}$$. $f$ is holomorphic everywhere except for $z_n=i\pi/2+2ni\pi$ for all $n$ in the integers. Let curve $C$ be closed positively oriented simple curve. \[ f(z_0)=\frac{1}{2i\pi}\int_C\frac{f(z)}{z-z_0}dz \], I want to find $z_0=i$, if it is possible.Thanks,

Cbarker1

 

[HOI]

I'm not sure what you mean by "find $z_0= i$". Do you mean find the value of the integral $\frac{1}{2\pi i}\int_C \frac{dz}{z- i}$?
Since that is equal to $f(z_0)= f(i)$, and $f(z)= \frac{z}{e^z- i}$ that is $\frac{i}{e^i- i}$. In general, $e^{ix}= cos(x)+ i sin(x)$ so $e^i= cos(1)+ i sin(1)$. $f(i)= \frac{i}{cos(1)+ i(sin(1)- 1}$. If you like you can "rationalize" the denominator by multiplying both numerator and denominator by $cos(1)- i(sin(1)- 1)$. That gives $\frac{1- sin(1)- i}{cos^2(1)+ sin^2(1)- 2sin(1)+ 1}= \frac{1- sin(1)- i}{2- 2sin(1)}$.

 

[cbarker1]

I'm not sure what you mean by "find $z_0= i$". Do you mean find the value of the integral $\frac{1}{2\pi i}\int_C \frac{dz}{z- i}$?
Since that is equal to $f(z_0)= f(i)$, and $f(z)= \frac{z}{e^z- i}$ that is $\frac{i}{e^i- i}$. In general, $e^{ix}= cos(x)+ i sin(x)$ so $e^i= cos(1)+ i sin(1)$. $f(i)= \frac{i}{cos(1)+ i(sin(1)- 1}$. If you like you can "rationalize" the denominator by multiplying both numerator and denominator by $cos(1)- i(sin(1)- 1)$. That gives $\frac{1- sin(1)- i}{cos^2(1)+ sin^2(1)- 2sin(1)+ 1}= \frac{1- sin(1)- i}{2- 2sin(1)}$.

I want to find the coefficient of Laurent series at the singularity point. Sorry for the confusion...