qr code java download (1 it )2 dt = u(t ) dt + v (t ) dt in Java

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(1 it )2 dt = u(t ) dt + v (t ) dt
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= (1 t ) dt i 2t dt
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Complex Integration
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These are elementary integrals that are easy to evaluate:
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(1 t 2 ) dt i 2t dt = t
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2 t3 8 2 it 2 = 2 4i = 4i 3 3 3 0
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ei 2 t dt
SOLUTION Using tools from elementary calculus we have
/4
ei 2 t dt =
/4 i i i 1 i 2t / 4 e = ei 2 t = ei / 2 + 2i 2 2 2 0 0
Now use Euler s formula: ei / 2 = cos( / 2) + i sin( / 2) = i And so the integral evaluates to
/4
i i 1+ i ei 2 t dt = (i ) + = 2 2 2
Properties of Complex Integrals
If f ( z ) is a function that depends on one real variable t such that f = u(t ) + iv (t ) then we can use theorems from the calculus of real variables to handle more complex integrals Suppose that = c + id is a complex constant You will recall from the calculus of real variables that we can pull a constant outside of an integral The same holds true here, where we have
f dt = (c + id )(u + iv ) dt = (c + id ) u dt + i (c + id ) v dt
a a a
(62)
Let g be another complex function depending on a single real variable such that g(t ) = r (t ) + is(t ) The integral of the sum or difference f g is
( f g) dt = f dt g dt
(63)
Complex Variables Demysti ed
Of course, we can also add the real and imaginary parts of the two functions:
( f g) dt = (u + iv ) (r + is) dt = (u r ) dt + i ( v s) dt
a a a
The product of two complex functions of a single real variable can be integrated as follows:
( fg) dt = (u + iv )(r + is) dt = (ur vs) dt + i ( vr + us) dt
a a a
(64)
As in the calculus of real variables, we can split up an interval a t b Suppose that a < c < b Then we can write
f (t ) dt = f (t )dt + f (t ) dt
(65)
Exchanging the limits of integration introduces a minus sign:
f (t ) dt = f (t )dt
(66)
The next example is somewhat contrived, since we could calculate the desired result easily, but it illustrates how the formulas could be applied and gives us practice calculating an integral of a complex function EXAMPLE 63 Given that 0 e t +it dt = e / 2 /2 [(1 i ) / 2], nd / 4 e t +it dt by calculating 0 e t +it dt SOLUTION The integral is easy to calculate We have
/2 /2 /4
/4
e t +it dt =
/4
e(1+i )t dt =
1 (1+i )t / 4 1 (1+i ) / 4 e = (e 1) 0 1+ i 1+ i
Euler s formula tells us that e(1+i ) / 4 = e / 4 ei / 4 = e / 4 (cos( / 4) + i sin( / 4)) ) A table of trigonometric functions can be consulted to learn that cos( / 4) = sin( / 4) = 2 1 = 2 2
And so
Complex Integration
e / 4 (1 + i ) 2
e(1+i ) / 4 =
Hence the integral is
/4
e t +it dt =
1 1 e / 4 e / 4 1 (e(1+i ) / 4 1) = (1 + i ) 1 = 2 1+ i 1+ i 2 1+ i
Writing the last term in standard form we obtain 1 1 1 i 1 i = = 1+ i 1+ i 1 i 2 Therefore:
/2
/4
e t +it dt =
e / 4 2
1 i 2
Now we use to write an expression that can be used to nd the desired integral: e t +it dt = e t +it dt = = =
/4
e t +it dt + e t +it dt
/2
/4 /4
e t +it dt e t +it dt
/2
/2
/4
e / 2 1 i e / 4 1 i e / 2 1 i e / 4 1 i + = 2 2 2 2 2 2 2 2 e / 2 e / 4 2 2
Contours in the Complex Plane
So far, we ve seen how to evaluate integrals of simple functions of a complex variable that were de ned in terms of a single real parameter we called t Now it s time to generalize and consider a more general case, where we just say we re integrating a function of a complex variable f ( z ), where z This can be done using a technique called contour integration The reason integrals of complex functions are done the way they are is that while an integral of a real-valued function is de ned on an interval of the line, an integral
Complex Variables Demysti ed
Figure 61 A curve (t ) is said to be simple if it does not cross itself
of a complex-valued function is de ned on a curve in the complex plane We say that a set of points in the complex plane z = ( x , y) is an arc if x = x (t ) and y = y(t ) are continuous functions of a real parameter t which ranges over some interval (ie, a t b) A complex number z can be written as z (t ) = x (t ) + iy(t ) De ne a curve as a continuous function (t ) that maps a closed interval a t b to the complex plane If the curve (t ) that de nes a given arc does not cross itself, which means that (t1 ) (t2 ) when t1 t2 , then we say that (t ) is a simple curve or Jordan arc A simple curve is illustrated in Fig 61 If the curve crosses over itself at any point, then it is not simple An example of this is shown in Fig 62 The curves in Figs 61 and 62 are open If (a) = (b), that is (t ) assumes the same value at the endpoints of the interval a t b, but at no other points, then we say that (t ) is a simple closed curve or closed contour This is shown in Fig 63 Formally, we say that a curve (t ) is a simple closed curve if (a) = (b) and (t ) is one-to-one
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