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This process can be continued For an arbitrary n, we obtain a second Cauchy s integral formula for the nth derivative of f (a): f ( n ) (a) = n! 2 i
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There are two facts you should come away with from Cauchy s integral formulas: If a function f (z) is known on a simple closed curve g, then that function is known at all points inside g Moreover, all of the functions derivatives can be found inside g If a function is analytic in a simply connected region of the complex plane, and hence has a rst derivative, all of its higher derivatives exist in that simply connected region Now we turn to a statement known as Cauchy s inequality This statement is related to Eq (72), which gives us an expression we can use to calculate the derivative of an analytic function in a simply connected region Consider a circle of radius r, which has the point z = a at its center, and suppose that f (z) is analytic on the circle and inside the circle Let M be a positive constant such that | f ( z) | M in the region | z a | < r Then f ( n ) (a) Mn ! rn (73)
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The next theorem, which is due to Liouville, tells us that an entire function cannot be bounded unless it is a constant This statement is called Liouville s theorem but it was rst proved by Cauchy So maybe we should call it the Cauchy-Liouville theorem In any case, it simply says that if f (z) is analytic and bounded in the entire complex plane, that is, f ( z) < M for some constant M, then f (z) is a constant
Residue Theory
Liouville s theorem implies the fundamental theorem of algebra Consider a polynomial with degree n 1 and coef cient an 0: P ( z) = a0 + a1z + a2 z 2 + + an z n
The fundamental theorem of algebra tells us that every polynomial P ( z) has at least one root The proof follows from Liouville s theorem and the use of a proof by contradiction Suppose that instead P ( z) 0 for all z Then f ( z) = 1 P( z)
is analytic throughout the complex plane and is bounded outside some circle z = r Moreover, the assumption that P ( z) 0 implies that f = 1/P is also bounded for z r Hence 1/P ( z) is bounded in the entire complex plane Using Liouville s theorem, 1/P ( z) must be a constant This is a contradiction, since P(z) = a0 + a1z + a2 z 2 + + an z n is clearly not constant Therefore P ( z) must have at least one root such that P ( z) = a0 + a1z + a2 z 2 + + an z n = 0 is satis ed Next we state the maximum modulus theorem and the minimum modulus theorem The maximum modulus theorem tells us the following Let f (z) be a complexvalued function which is analytic inside and on a simple closed curve g If f (z) is not a constant, then the maximum value of f ( z) is found on the curve g Now we state the minimum modulus theorem Assume once again that f (z) is a complex-valued function which is analytic inside and on a simple closed curve g If f ( z) 0 inside g, then f ( z) assumes its minimum value on the curve g The next theorem is the deformation of path theorem Consider a domain D in the complex plane, and two curves in D we call 1 and 2 We suppose that 1 is larger than or lies outside of 2 , and that 1 can be deformed into 2 without leaving the domain D [that is, we can shrink the rst curve down to the second one without crossing any holes or discontinuities in the domain (Fig 71)] If f (z) is analytic in D then
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