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Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer.

eiz(z2+4)2atz=2i

Short Answer

Expert verified

The residue of the function at z=2iis R(2i)=3i32e-2

Step by step solution

01

Determine the formula for the higher order poles 

Residue of a function at higher order poles is given as follows:

R[f(z)]=1(n-1)!limz→zkdn-1dzn-1(z-zk)nf(z)

02

Determine the residue of the higher order pole:

Consider the given function as:

f(z)=eiz(z2+4)2

Resolve the function as:

f(z)=eiz(z2+4)2=eiz(z2+2i)2(z-2i)2

At z = 2i

The function has a pole of order n = 2 at z = 2i and the residue of higher order pole is given by:

R(z0)=1(n-1)!limz→z0dn-1dzn-1(z-z0)f(z) ……. (1)

03

 Determine the residue of the higher order function as:

Substitute the values in equation (1) and solve as:

R(2i)=1(n-1)!limz→0ddzeiz(z+2i)2=ieiz(z+2i)2-2eiz(z+2i)(z+2i)4z=2i=ieiz(z+2i)-2eiz(z+2i)3z=2i

Solve further as:

R(2i)=ieiz(z+2i)-2eiz(z+2i)3z=2i=ieiz(4i)-e-2×2(4i)3=3i32e-2

Therefore, the residue of a function at z = 2i is R(2i)=3i32e-2.

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