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qr code java download f ( z) dz = f ( z) dz in Java
f ( z) dz = f ( z) dz QRCode Generation In Java Using Barcode creator for Java Control to generate, create QR image in Java applications. Quick Response Code Scanner In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. (74) Paint Barcode In Java Using Barcode drawer for Java Control to generate, create barcode image in Java applications. Reading Barcode In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. Next, we state Gauss mean value theorem Consider a circle g of radius r centered at the point a Let f (z) be a function, which is analytic on and inside g The mean value of f (z) on g is given by f (a): f (a) = 1 2 Denso QR Bar Code Drawer In Visual C#.NET Using Barcode generation for .NET framework Control to generate, create QR Code ISO/IEC18004 image in VS .NET applications. QR Code Maker In .NET Using Barcode printer for ASP.NET Control to generate, create QR Code JIS X 0510 image in ASP.NET applications. 2 0 QR Code Maker In Visual Studio .NET Using Barcode drawer for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in .NET applications. QR Code JIS X 0510 Generator In VB.NET Using Barcode generation for VS .NET Control to generate, create QR Code image in Visual Studio .NET applications. f (a + rei ) d
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Data Matrix ECC200 Drawer In Visual C# Using Barcode maker for .NET framework Control to generate, create Data Matrix image in .NET framework applications. Code 128C Drawer In VS .NET Using Barcode encoder for Reporting Service Control to generate, create Code 128A image in Reporting Service applications. Once again, let f (z) be a function, which is analytic on and inside a simple, closed curve g Now assume that f (z) has a nite number of poles inside g If M is the number of zeros of f (z) inside g and N is the number of poles inside g, the argument theorem states that 1 2 i f ( z) dz = M N f ( z) (76) Next is a statement of Rouche s theorem Let f (z) and g (z) be two functions, which are analytic inside and on a simple closed curve g If g( z) f ( z) on , then f ( z) + g( z) and f ( z) have the same number of zeros inside g Finally, we end our whirlwind tour of theorems and results related to the Cauchy s integral formula with a statement of Poisson s integral formula for a circle This expresses the value of a harmonic function inside of a circle in terms of its values on the boundary Let f (z) be analytic inside and on the circle g, centered at the origin with radius R Suppose that z = rei is any point inside g Then f ( z) = 1 2 2 0 ( R 2 r 2 ) f (Rei ) d R 2 2 Rr cos( ) + r 2
(77) EXAMPLE 71 This example illustrates the solution of Laplace s equation on a disk First show that u (r , ) = a0 + an r n cos n + bn r n sin n n =1 Residue Theory
is the solution of Laplace s equation on the disc 0 r 1 with Dirichlet boundary conditions: 1 u 1 2u =0 r + r r r r 2 r 2 0 < r < 1, 0 2 u (1, ) = f ( ), u (r , ) bounded as r 0 Show the coef cients in the series expansion are given by a0 = 1 2 2 0 f ( ) d
an =
2 0 f ( ) cos n d
bn =
2 0 f ( )sin n d
Use the result to deduce Poisson s integral formula for a circle of radius one: u (r , ) = 1 2 2 0 1 r2 f ( ) d 1 2r cos( ) + r 2
SOLUTION We try separation of variables Let u (r , ) = R(r ) ( ) Then it follows that u R = ( ) r r 2u 2 R = ( ) r 2 r 2 2u 2 = R( r ) 2 2 The statement of the problem tells us that 2u 1 u 1 2u =0 + + r 2 r r r 2 2 Hence 0= 2 2 R 1 R 1 ( ) + ( ) + 2 R(r ) 2 r r r r 2 We divide every term in this expression by u (r , ) = R(r ) ( ) This allows us to write r 2 2 R r R 1 2 = + 2 R r 2 R r Complex Variables Demysti ed
The lefthand side and the righthand side are functions of r only and only, respectively Therefore they can be equal only if they are both equal to a constant We call this constant n 2 Then we have the equation in : 1 d 2 = n2 d 2 d 2 + n 2 = 0 d 2 Note that partial derivatives can be replaced by ordinary derivatives at this point, since each equation involves one variable only This familiar differential equation has solution given by ( ) = an cos n + bn sin n Now, turning to the equation in r, we have r 2 d 2 R r dR + = n2 2 R dr R dr r2 d2R dR +r n2 R = 0 2 dr dr You should also be familiar with this equation from the study of ordinary differential equations It has solution R(r ) = cn r n + c n r n The total solution, by assumption is the product of both solutions, that is, u (r , ) = R(r ) ( ) So we have u (r , ) = (cn r n + c n r n )(an cos n + bn sin n ) The condition that u (r , ) is bounded as r 0 imposes a requirement that the constant c n = 0 since c n as r 0 rn Therefore, we take u (r , ) = (cn r n )(an cos n + bn sin n ) We can just absorb the constant cn into the other constants, and still designate them by the same letters Then u (r , ) = r n (an cos n + bn sin n )

