# qr code reader java app Residue Theory in Java Creation QR Code JIS X 0510 in Java Residue Theory

Residue Theory
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And so, shows that the residue corresponding to the pole at z = exp(i 3 /4) is given by z = ei 3 / 4 (1/ 4)ei / 4 Hence
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residues = 4 e
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1 ei 3 / 4 4
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1 = (cos / 4 + i sin / 4 + cos 3 / 4 + i sin 3 / 4) 4 i 1 1 1 = (1 + i 1 + i ) = 2i = 4 2 4 2 2 2 Therefore the integral evaluates to i I = 2 i residues in upper half plane = 2 i = 2 2 2 EXAMPLE 712 Compute [(cos x )/( x 2 2 x + 2)]dx SOLUTION We can compute this integral by considering
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eiz dz = z 2 2z + 2
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eiz dz [ z (1 + i )][ z (1 i )]
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The root z = 1 + i lies in the upper half plane, while the root z = 1 i lies in the lower half plane We choose a contour which is a semicircle in the upper half plane, enclosing the rst root This is illustrated in Fig 78 The residue is given by eiz lim [ z (1 + i ) ] z i +1 [ z (1 + i)][ z (1 i)] = lim = eiz z i +1 z (1 i ) e 1ei 2i
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Complex Variables Demysti ed
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y C1 1+i C2 R 0 1 i R x
Figure 78 The contour used in Example 712
Therefore we have
e 1ei i eiz dz = 2 i = e z 2 2z + 2 2i e
But, using Euler s identity, we have ei = cos 1 + i sin 1 And so
Now, we have
eiz dz = cos(1) + i sin(1) 2 z 2z + 2 e e
eix dx eiz dz + 2 R x 2 2 x + 2 C z 2 z + 2 1 = =
R cos xdx sin xdx eiz dz + i + 2 R x2 2x + 2 R x2 2x + 2 z 2z + 2 C1 R
cos(1) + i sin(1) e e
Now we let R By Jordan s lemma: eiz dz =0 2 C1 z 2 z + 2
So we have
Residue Theory
cos xdx sin xdx + i 2 = cos(1) + i sin(1) x 2 x + 2 x 2 x + 2 e e 2
Equating real and imaginary parts gives the result we are looking for:
cos xdx = cos(1) x 2x + 2 e
Summary
By computing the Laurent expansion of an analytic function in a region containing one or more singularities, we were able to arrive at the residue theorem which can be used to calculate a wide variety of integrals This includes integrals of complex functions, but the residue theorem can also be used to calculate certain classes of integrals involving functions of a real variable
Quiz
1 Compute
sinh z dz z3 sinh z dz z4
2 Compute
3 Find the principal part of f ( z ) =
1 (1 + z 3 )2
sin z 5 z z + 2 sin z 5 What are the singularities and residues of 2 z ( z ) d 6 Evaluate 2 0 24 6 sin 4 What are the singular points and residues of
Complex Variables Demysti ed
dx 8 Use the residue theorem to compute lim R R 1 + x 2 x2 + 3 9 Compute ( x 2 + 1)( x 2 + 4) dx cos x 10 Compute dx 1 + x 2
7 Using the technique outlined in Example 79, compute
sin 2 x dx x2
More Complex Integration and the Laplace Transform
In this chapter, we consider a few more integrals that can be evaluated using the residue theorem and then consider the Laplace transform
Contour Integration Continued
Consider the Fesnel integrals which are given by
cos(t 2 ) dt = sin(t 2 ) dt =
2 2
(81)
Complex Variables Demysti ed
III p/4 I
II C x
Figure 81 The contour C used to evaluate the integrals in Eq (81)
These integrals can be evaluated by considering a wedge in the rst quadrant of the complex plane with angle = /4 This is illustrated in Fig 81 The three legs along the contour have been denoted by I, II, and III We consider the analytic function f (z) = e z
(82)
If we integrate the function in Eq (82) around C, we will nd it to be zero This can be done using the residue theorem which tells us that
z2
dz = 2 i enclosed residues
This function has no singularities, a fact we can verify explicitly by writing down its series representation: e z = 1 z 2 +
1 4 1 6 z z + 2! 3!
Therefore
enclosed residues
And so C e z dz = 0 Now let s try a different approach First we break up the integral into separate integrals along each of the curves I, II, and III:
z2
dz = e z dz + e z dz + e z dz
2 2 2
More Complex Integration
We take the radius of the wedge to be xed (for now) at r = R Now write down the polar representation: z = rei and dz = dr (ei ) + irei d These quantities will assume different values on each leg of the contour First consider curve I, which lies on the x axis Along curve I, d = 0, = 0, z = r , dz = dr Curve II is a circular path at radius R So while varies, r is xed So along curve II, r = R, dr = 0, dz = iRei d Finally, along curve III, is once again xed like it was on curve I But this time, = / 4, d = 0, dz = ei / 4 dr Using these results the integral becomes
z2