vba code for barcode in excel Figure 62 A curve which crosses itself at one or more points is not simple in Java

Printing QR in Java Figure 62 A curve which crosses itself at one or more points is not simple

Figure 62 A curve which crosses itself at one or more points is not simple
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Complex Integration
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Figure 63 A simple, closed curve
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When using contour integration, the sense or direction in which the curve is traversed is important To understand this, we consider a simple example, the unit circle centered about the origin For example, consider z = ei where 0 2 If you put in some values as q ranges over the given interval increasing from 0, you will note that the points sweep out the circle in the counterclockwise direction To see this, write the points in the complex plane as z = ( x , y) Let s plug in a few points:
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= 0 z = ei 0 = cos 0 + i sin 0 = (1, 0)
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1 1 = / 4 z = ei / 4 = cos( / 4) + i sin( / 4) = , 2 2
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= / 2 z = ei / 2 = cos( / 2) + i sin( / 2) = (0,1) = z = ei = cos + i sin = ( 1, 0)
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Following the curve in the counter-clockwise direction can be said to be in the positive sense since it moves with increasing angle When drawing a contour, we use an arrow to indicate the directional sense we are using to move around it This is illustrated in Fig 64 If we move around the curve in the opposite direction, which is clockwise, we ll call that negative because we will be moving opposite to the direction of increasing angles Now consider the function: z = e i
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Figure 64 A closed contour traversed in the positive sense, which is counter-clockwise We say that this is in the positive sense because the curve is traversed in the direction of increasing angle q in the complex plane
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This also describes the unit circle, but we are traversing the circle in the counterclockwise direction Notice that
= 0 z = e i 0 = cos 0 i sin 0 = (1, 0)
1 1 = / 4 z = e i / 4 = cos( / 4) i sin( / 4) = , 2 2
= / 2 z = e i / 2 = cos( / 2) i sin( / 2) = (0, 1) = z = e i = cos i sin = ( 1, 0)
The case of traversing a circle in the clockwise or negative direction is illustrated in Fig 65
Complex Line Integrals
In this section, we will formalize what we ve done so far with integration a bit First let s review important properties a function must have so that we can integrate it DEFINITION: CONTINUOUSLY DIFFERENTIABLE FUNCTION Let a function f (t ) map the interval a t b to the real numbers Formally, we write f : [a, b] We say that f (t ) is continuously differentiable over this interval, 1 which we indicate by writing f C ([a, b]) if the following conditions are met: The derivative df/dt exists on the open interval a < t < b The derivative df/dt has a continuous extension to a t b
Complex Integration
Figure 65 Traversing the contour in the negative or clockwise direction
This allows us to utilize the fundamental theorem of calculus This tells us that
f (t ) dt = f (b) f (a)
(67)
Now we will extend this to curves in the complex plane Suppose that a curve (t ) = f (t ) + ig(t ) DEFINITION: CONTINUOUSLY DIFFERENTIABLE CURVE Let (t ) be a curve, which maps the closed interval a t b to the complex plane We say that (t ) is continuous on a t b if f (t ) and g(t ) are both continuous on a t b If f (t ) and g(t ) are both continuously differentiable functions on a t b, then the curve (t ) is continuously differentiable This is indicated by writing C 1 ([a, b]) If (t ) is continuously differentiable and (t ) = f (t ) + ig(t ), then the derivative is given by dg d df = +i dt dt dt (68)
We ve already seen that we can write the integral of w(t ) = u(t ) + iv (t ) as b u(t )dt + i b v (t )dt If the curve (t ) = f (t ) + ig(t ) is continuously differentiable, a a then we can write what might be called the fundamental theorem of calculus for curves in the complex plane:
(t ) dt = (b) (a)
(69)
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