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Chapter 14: Q24P (page 687)

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.

1-cos2zz3atz=0

Short Answer

Expert verified

Hence, the residue of the function at z0=0 is 2.

Step by step solution

01

Given information

The given function is: fz=1-cos2zz3.

02

Residue Theorem

If z0is an isolated singular point of f(z). Then the integration of the function within any closed curve C is given by:

鈭�cfzdz=b1路2蟺i

Here, b1 is the residue.

03

Step 3:Find the Residue

Since, z=0is a pole of higher order of 3.

So, the residue of this type of function is given by:

Resz=z0fz=1n-1!limz鈫�z0dn-1dzn-1fz路z-z0n

According to the question, we have:z0=0andn=3

Now, the residue atz0=0will be:

Resz=0fz=13-1!limz鈫�0d3-1dz3-11-cos2zz3路z-03=12!limz鈫�0d2dz21-cos2z=12!limz鈫�04cos2z=12路4=2

Hence, the residue of the function at z0=0 is 2.

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Using the definition (2.1) of , show that the following familiar formulas hold. Hint : Use the same methods as for functions of a real variable.

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In equation (7.18), let u (x) be an even function and 蠀(x)be an odd function.

  1. If f(x)=u(x)+i蠀(x), show that these conditions are equivalent to the equationf*(x)=f(-x) .
  2. Show that

蟺u(a)=PV鈭�0鈭�2x蠀(x)x2-a2鈥�dx,鈥�鈥�鈥�鈥�鈥�鈥�蟺蠀(a)=-PV鈭�0鈭�2au(x)x2-a2鈥�dx

These are Kramers-Kroning relations. Hint: To find u(a), write the integral for u(a) in (7.18) as an integral from -鈭�to 0 plus an integral from 0 to 鈭�. Then in the to integral -鈭�to 0, replace x by -x to get an integral from 0 to鈭� , and userole="math" localid="1664350095623" 蠀(-x)=-蠀(x) . Add the two to integrals and simplify. Similarly findrole="math" localid="1664350005594" 蠀(a) .
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