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Complex Theorem for Continuous Anti-derivative

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Although Corollary 2 does not apply to the function 1 / (z — z_0) in the plane punctured at z_0, Theorem 6 can be used as follows to show that
(see attached 1)
for any circle C traversed once in the positive direction surrounding the point z_0. Introduce a horizontal branch cut from z_0 to infinity as in Fig. 4.25. In the resulting "slit plane" the function 1 / (z — z_0) has the anti-derivative Log (z — z_0). Apply Theorem 6 to compute the integral along the portion of C from alpha to beta as indicated in Fig. 4.25. Now let alpha and beta approach the point t on the cut to evaluate the given integral over all of C.
(see attached 1)
Theorem 6. Suppose that the function f (z) is continuous in a domain D and has an anti-derivative F(z) throughout D; i.e., dF(z)Idz = f (z) for each z in D. Then for any contour gamma lying in D, with initial point z_I and terminal point z_T, we have
(see attached 2).

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Solution Summary

The solution assists with using the complex theorem for continuous anti-derivative and computing the integrals.

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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