Question

the residue of 1/(z^2+1) at z=i is​

Answer 1

Answer:

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Answer 2

Answer:

The residue of  f(z) = 1/(z²+1) at z = i is 1/(z+i).

Step-by-step explanation:

Residue of nth order :

• The residue of a function at the pole 'a' with nth order

              Res f(a) =    $\frac{1}{(n-1)!} \frac{d^{n-1} }{dx^{n-1} } (z-a)^{n}f(z)$  

 Given function is  f(z) = 1/(z²+1).

    we need to find the residue at z = i        

            f(z) = 1/(z²+1)

                  = 1/(z²-(-1))

                  = 1/(z²-i²)          

            f(z)  = 1/(z+i)(z-i)              

        At z = i and z = -i the function f(z) is zero.      

    So, the Poles of the function f(z) are i and -i are of order 1.  

    We need to find the residue of function at z = i of order n = 1.

           Res f(i) =   $\frac{1}{(1-1)!} \frac{d^{1-1} }{dx^{1-1} } (z-i)^{1}\frac{1}{(z-i)(z+i)}$  

                       =  (1/1) (z-i)/[(z-i)(z+i)]

           Res f(i) = 1/(z+i)

         Hence , the residue of  f(z) = 1/(z²+1) at z = i is 1/(z+i).        

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