# Complex integration using series expansion of analytic functions

I want to check my answer:

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.

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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)

#### Solution Preview

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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.

Yes, you ...

#### Solution Summary

The complex integration using series expansion of analytic functions.