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this module covers the residue theorem, removable and isolated singularities, some new developments from Cauchy's theorem, and related exercises.

The first result we present in this section is a natural extension of [link] . However, as we shall see, its consequences for computing contour integrals can hardly be overstated.

Let S be a piecewise smooth geometric set whose boundary C S has finite length. Suppose c 1 , ... , c n are distinct points in the interior S 0 of S , and that r 1 , ... , r n are positive numbers such that the closed disks { B ¯ r k ( c k ) } are contained in S 0 and pairwise disjoint. Suppose f is continuous on S B r k ( c k ) , i.e., at each point of S that is not in any of the open disks B r k ( c k ) , and that f is differentiable on S 0 B ¯ r k ( c k ) , i.e., at each point of S 0 that is not in any of the closed disks B ¯ r k ( c k ) . Write C k for the circle that is the boundary of the closed disk B ¯ r k ( c k ) . Then

C S f ( ζ ) d ζ = k = 1 n C k f ( ζ ) d ζ .

This is just a special case of part (d) of [link] .

Let f be continuous on the punctured disk B ' ¯ r ( c ) , analytic at each point z in B r ' ( c ) , and suppose f is undefined at the central point c . Such points c are called isolated singularities of f , and we wish now to classify these kinds of points. Here is the first kind:

A complex number c is called a removable singularity of an analytic function f if there exists an r > 0 such that f is continuous on the punctured disk B ' ¯ r ( c ) , analytic at each point in B r ' ( c ) , and lim z c f ( z ) exists.

  1. Define f ( z ) = sin z / z for all z 0 . Show that 0 is a removable singularity of f .
  2. For z c , define f ( z ) = ( 1 - cos ( z - c ) ) / ( z - c ) . Show that c is a removable singularity of f .
  3. For z c , define f ( z ) = ( 1 - cos ( z - c ) ) / ( z - c ) 2 . Show that c is still a removable singularity of f .
  4. Let g be an analytic function on B r ( c ) , and set f ( z ) = ( g ( z ) - g ( c ) ) / ( z - c ) for all z B r ' ( c ) . Show that c is a removable singularity of f .

The following theorem provides a good explanation for the term “removable singularity.” The idea is that this is not a “true” singularity; it's just thatfor some reason the natural definition of f at c has not yet been made.

Let f be continuous on the punctured disk B ¯ r ' ( c ) and differentiable at each point of the open punctured disk B r ' ( c ) , and assume that c is a removable singularity of f . Define f ˜ by f ˜ ( z ) = f ( z ) for all z B r ' ( c ) , and f ˜ ( c ) = lim z c f ( z ) . Then

  1.   f ˜ is analytic on the entire open disk B r ( c ) , whence
    f ( z ) = k = 0 c k ( z - c ) k
    for all z B r ' ( c ) .
  2. For any piecewise smooth geometric set S B r ( c ) , whose boundary C S has finite length, and for which c S 0 ,
    C S f ( ζ ) d ζ = 0 .

As in part (a) of [link] , define F on B r ( c ) by

F ( z ) = 1 2 π i C r f ( ζ ) ζ - z d ζ .

Then, by that exercise, F is analytic on B r ( c ) . We show next that F ( z ) = f ˜ ( z ) on B r ( c ) , and this will complete the proof of part (1).

Let z be a point in B r ( c ) that is not equal to c , and let ϵ > 0 be given. Choose δ > 0 such that δ < | z - c | / 2 and such that | f ˜ ( ζ ) - f ˜ ( c ) | < ϵ if | ζ - c | < δ . Then, using part (c) of [link] , we have that

f ˜ ( z ) = f ( z ) = 1 2 π i C r f ( ζ ) ζ - z d ζ - 1 2 π i C δ f ( ζ ) ζ - z d ζ = F ( z ) - 1 2 π i C δ f ( ζ ) - f ˜ ( c ) ζ - z d ζ - 1 2 π i C δ f ˜ ( c ) ζ - z d ζ = F ( z ) - 1 2 π i C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ ,

where the last equality holds because the function f ˜ ( c ) / ( ζ - z ) is an analytic function of ζ on the disk B δ ( c ) , and hence the integral is 0 by [link] . So,

| f ˜ ( z ) - F ( z ) | = | 1 2 π i C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ | 1 2 π C δ | f ˜ ( ζ ) - f ˜ ( c ) | | ζ - z | d s 1 2 π C δ ϵ δ / 2 d s = 2 ϵ δ × δ = 2 ϵ .

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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I'm not sure why it wrote it the other way
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Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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