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qr code reader java app Complex Variables Demysti ed in Java
Complex Variables Demysti ed Create Quick Response Code In Java Using Barcode printer for Java Control to generate, create QR image in Java applications. Recognize Quick Response Code In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. a B C D x
Painting Barcode In Java Using Barcode maker for Java Control to generate, create bar code image in Java applications. Decode Barcode In Java Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications. Figure 915 To map a vertical strip to the upper half plane, we utilized the transformation w = sin z / a It maps to the region shown in Fig 916 Print QR Code ISO/IEC18004 In C#.NET Using Barcode generator for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications. QRCode Drawer In .NET Framework Using Barcode maker for ASP.NET Control to generate, create QR Code 2d barcode image in ASP.NET applications. Rules of Thumb
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Barcode Encoder In None Using Barcode drawer for Font Control to generate, create barcode image in Font applications. Create ECC200 In ObjectiveC Using Barcode drawer for iPhone Control to generate, create Data Matrix 2d barcode image in iPhone applications. The transformation w = sin( z / a) maps the region in Fig 915 to the region in Fig 916 with corresponding points indicated Painting Code 128 Code Set C In VB.NET Using Barcode generator for Visual Studio .NET Control to generate, create Code 128 image in Visual Studio .NET applications. Barcode Generator In None Using Barcode creation for Software Control to generate, create barcode image in Software applications. Mapping and Transformations
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Figure 917 For review, a transformation of the form w = az expands the region of interest if  a  >1 Now suppose that w = (1/ 2) z This shrinks the square, as shown in Fig 918 To rotate the region by an angle , we use a transformation of the form w = ei z (94) For our square, this rotates the square by in the counterclockwise direction assuming that > 0 This is illustrated in Fig 919 M bius Transformations
In this section, we consider a transformation of the type: Tz = az + b cz + d ad bc 0 (95) w = 1/2z
Figure 918 We shrink the square by the transformation w = az when  a  <1
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w = eifz f
Figure 919 A rotation is implemented with a transformation of the form w = ei z
This type of transformation goes under the various names bilinear transformation, fractional transformation, or M bius transformation The transformation shown in Eq (95) is actually a composition of three different transformations These are Dilation, which can be written as the linear transformation az Translation, which is written as z + b Reciprocation, which is the transformation 1/z The requirement that ad bc 0 is based on the following The derivative of Eq (95) is given by (Tz ) = a(cz + d ) c(az + b) (cz + d )2 Evaluating this at z = 0 we have (Tz ) (0) = ad bc (d ) 2 This tells us that the transformation in Eq (95) will be a constant unless ad bc 0 A transformation of the type in Eq (95) maps circles in the z plane to circles in the w plane Straight lines are also mapped into straight lines Now suppose that z0 , z1 , z 2 , and z3 are four distinct points in the complex plane The cross ratio is given by ( z3 z0 )( z1 z 2 ) ( z1 z0 )( z3 z2 ) (96) Mapping and Transformations
The cross ratio is invariant under a M bius transformation That is if z j w j under a M bius transformation, then ( z3 z0 )( z1 z 2 ) ( w w0 )( w1 w2 ) 3 ( z1 z0 )( z3 z2 ) ( w1 w0 )( w3 w2 ) There are a few M bius transformations of interest Let a be a complex number with  a  < 1 and suppose that  k  = 1 Then w=k z a 1 az (97) maps the unit disk from the z plane to the unit disk in the w plane Now let a be a complex number with the requirement that Im(a) > 0 The transformation w=k z a z a (98) maps the upper half of the z plane to the unit disk in the w plane Notice that when z is purely real,  w  =  k  = 1 EXAMPLE 94 Consider a disk of radius r = 2 centered at the point z = 1 + i Find a transformation that will take this to the entire complex plane with a hole of radius 1/2 centered at the origin SOLUTION Since these transformations are linear, we can do this by taking multiple transformations in succession First we illustrate what we re starting with, a disk of radius r = 2 centered at the point z = 1 i This is shown in Fig 920

