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Complex Variables Demysti ed
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This result can be extended further We consider a domain D in the complex plane and a curve (t ) which maps a real, closed interval a t b into D If there is a continuously differentiable function h, which maps D into the real numbers, then the integral along the curve is given by
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h df h dg ) x (t ) dt + y (t ) dt dt = h( (b)) h( (a))
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(610)
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DEFINITION: COMPLEX LINE INTEGRAL OR CONTOUR INTEGRAL Now suppose that the curve (t ) is a simple closed curve Then the complex line integral of a function F ( z ) of a complex variable is written as
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F (z ) dz =
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F ( (t ))
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d dt dt
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The integral in Eq (611) is known as a contour integral EXAMPLE 64 Suppose that 0 t 1, f ( z ) = z and we integrate along the curve (t ) = 1 + (i 1)t Calculate f ( z )dz SOLUTION This can be done by using Eq (611) Given that (t ) = 1 + (i 1)t , we see that d = i 1 dt We also have that f ( z ) = f ( (t )) = z = 1 + (i 1)t and so
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F ( (t ))
1 d dt = (1 + (i 1)t )(i 1) dt 0 dt
= (i 1) (1 + (i 1)t ) dt
t2 1 = (i 1) t + (i 1) 2 0 i + 1 = (i 1) = 1 2
Complex Integration
EXAMPLE 65 Suppose that f ( z ) = z 2 + 1 Integrate f ( z ) around the unit circle SOLUTION We can integrate around the unit circle by de ning the curve:
(t ) = eit
The interval mapped by this curve to the complex plane is 0 t 2 We nd that the derivative of the curve is d d = (eit ) = ieit dt dt Using Eq (611) we have
f ( z ) dz = f ( (t ))
a 2 0
2 d d dt = ( 2 + 1) dt 0 dt dt
= ((eit )2 + 1)(ieit ) dt = i (ei 3 t + eit ) dt
0 2
2 1 = ei 3 t + eit 3 0 1 = (ei 6 1) + (ei 2 1) = 0 3 This result follows since for any even m, eim = cos(m ) + i sin(m ) = 1 + i 0 = 1
The Cauchy-Goursat Theorem
Now let s take a turn that we re going to use to develop the groundwork for residue theory, the topic of the next chapter First let s begin by looking at complex integration once again We ll dispense with the parameter t and instead focus on functions of x and y So we have w = f ( z ) = u( x , y) + iv ( x , y)
With z = x + iy , then
Complex Variables Demysti ed
dz = dx + idy
(612)
So we can write the integral of a complex function along a curve in the following way:
f (z ) dz = (u + iv)(dx + idy) = udx vdy + i vdx + udy
(613)
With this in hand, we can de ne the fundamental theorem of calculus for a function of a complex variable as follows Suppose that f ( z ) has an antiderivative That is: f (z) = dF dz
The fundamental theorem of calculus then becomes
f (z ) dz = dz dz = F (z ) a = F (b) F (a)
(614)
To prove this result, we use F ( z ) = U + iV We are assuming that f and F are analytic Now, using the results of Chap 3 we know that f (z) = and so U dF U V V i = = +i y dz x x y
f (z )dz = dz dz
V V U U dy dx + dy + i dx + y x y x
But since U = U ( x , y) using the chain rule we know that dU = U U dx + dy y x
Complex Integration
and similarly for dV Hence
f ( z ) dz = dz dz
= dU + i dV
z=b b + iV z=a a
= F (b) F (a) The fundamental theorem of calculus allows us to evaluate many integrals in the usual way EXAMPLE 66
n Find f ( z )dz when f ( z ) = z for the case of z (= a) z (= b), and when the curve is closed, that is when a = b
SOLUTION The fundamental theorem allows us to evaluate the integral in the same way we would in the calculus of real variables We have