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f ( z) dz = f ( z) dz
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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
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f (a + rei ) d
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g1 Can have hold inside second curve
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Figure 71 A graphic illustration of the deformation of path theorem
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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)
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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
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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
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The left-hand side and the right-hand 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 )
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