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Complex integration using series expansion of analytic functions

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Evaluate the following integrals:

integral over gamma for (sin z)/z dz, given that gamma(t) = e^(it) , 0=<t=<2pi
( e here is the exponential function)

My work:

sin z = z - z^3/3! + z^5/5! + ... + (-1)^n (z^(2n-1))/(2n-1)! + ...

divide by z we get

(sin z)/z = 1 - z^2/3! + z^4/5! - ... + (-1)^n ( z^(2n-2)/(2n-1)!) +...

Then since we don't have 1/z , when we integrate we get zero for all
So the integral is 0.

------------------------------------------------------

Second one is to evaluate the following integral:

integral over gamma of dz/(z-1/2)^n, n is positive, gamma(t) = 1/2 + e^(it)

My work:

I will rewrite the integral as a function of gamma(t)
gamma ' (t) = d/dt gamma(t) = i e^it that will replace the dz
Then ( 1/2 + e^it - 1/2)^n that's in the numirator, and it gives (e^int) in the numirator
Also I will evaluate the gamma(t) at the bounday values 0 =<t=<2 pi
so my integral will be from 1/2 to 1/2+ e^i2pi

Now put all that together, we get:
integral of (i e^it)/(e^int) dt,
I will factor out the i from the integral and will work with e^it/e^int = e^i(1-n)t

Now the series expansion for that term is
e^i(1-n)t = 1 + 1 + (-i t)^2/2! + (-2i t)^3/3! + ... + ( (1-n)it) )^n/n! + ...
all integrals will be zero except for the 1+1 = 2 we integrate it we get
integral from 1/2 to 1/2+e^i 2 pi of i(2) dt
we get 2i e^i2pi
Am I correct? please let me know. Thanks. ( please reply as soon as possible)

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The complex integration using series expansion of analytic functions.

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Please see the attached file.

Evaluate the following integrals:

integral over gamma for (sin z)/z dz, given that gamma(t) = e^(it) , 0=<t=<2pi
( e here is the exponential function)

My work:

sin z = z - z^3/3! + z^5/5! + ... + (-1)^n (z^(2n-1))/(2n-1)! + ...

divide by z we get

(sin z)/z = 1 - z^2/3! + z^4/5! - ... + (-1)^n ( z^(2n-2)/(2n-1)!) +...

Then since we don't have 1/z , when we integrate we get zero for all
So the integral is 0.

Yes, you ...

Solution provided by:
Changping Wang, MA
Education
  • 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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