java barcode scanner api In Example 94, we start with a region de ned by a disk centered at z = 1 + i in Java

Generate Denso QR Bar Code in Java In Example 94, we start with a region de ned by a disk centered at z = 1 + i

In Example 94, we start with a region de ned by a disk centered at z = 1 + i
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Figure 921 A disk at the origin is obtained from the disk shown in Fig 920 via the transformation Z = z z0 = z + 1 i We denote this the Z plane
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The rst step is to move the disk to the origin We do this using Z = z z0 = z + 1 i The result is the disk shown in Fig 921 Now we want to transform the disk shown in Fig 921 so that the region of de nition is the entire complex plane minus a hole where the disk was We do this using an inverse transformation: w= The result is shown in Fig 922 1 Z
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Figure 922 The transformation 1/Z changes the region to the entire complex plane with a hole punched out in the middle The radius of the hole is 1/r if the radius of the disk we started with was Z = rei In our example, r = 2 so the hole here has a radius = 1/ 2
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The complete transformation in this example can be written as w= 1 z +1 i
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This is a M bius transformation as in Eq (95) with a = 0, b = 1, c = 1, and d = 1 + i EXAMPLE 95 Construct a M bius transformation that maps the unit disk to the left half plane Re( z ) < 0 and one that maps the unit disk to the right half plane Re( z ) > 0 SOLUTION The rst transformation we want to consider is illustrated in Fig 923 First we consider the boundary of the disk, which is the unit circle, that is the set of points | z | = 1 For the transformation shown in Fig 923 to work, we must map the points on the unit circle to the imaginary axis In the form of a M bius transformation, the mapping will be of the form Tz = az + b cz + d
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This transformation has a pole located at the point z = d /c We are free to pick a point on the unit circle to map to the pole, so we choose z = 1 With this choice we have the freedom to x c and d, so we choose c = 1, d = 1 So Tz = az + b z 1
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Figure 923 We want to map the unit disk to the left half plane
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Second, we need to pick a point z on the unit circle such that Tz = 0 We have already used the point z = 1, so we choose z = 1 This forces us to take a = b since Tz = 0 when z = 1 We can choose a = 1 giving us the transformation Tz = z +1 z 1
Now we can see how the transformation maps the rest of the unit disk The simplest point to check is the point z = 0 at the center of the disk We have Tz(0) = 0 +1 = 1 0 1
So the transformation maps the center of the disk z = 0 to the point z = 1 which is in the left half plane Hence this is the transformation that we want To transform the unit disk to the right half plane instead, it turns out we are almost there All we have to do is rotate the transformed region shown in Fig 922 The angle that is required is , and a rotation is implemented by multiplication by ei So the transformation that takes the unit disk to the right half plane is given by Tz = ei z +1 z + 1 = z 1 z 1
EXAMPLE 96 Consider a mapping that will transform the unit disk into the upper half plane
SOLUTION We again seek a transformation of the form Tz = az + b cz + d
This time we choose to map the point z = 1 onto the pole at z = d /c If we choose c = 1, then d = 1 as well and the transformation is given by Tz = az + b z +1
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