Evaluate integral around a closed contour of f(z) dz f(z)=(sin z)/z

[blueyellow]

Homework Statement

evaluate integral around a closed contour (C) of f(z) dz, where C is the unit circle centred at the origin and f(z) is (sin z)/z

Homework Equations

The Attempt at a Solution

well, the textbooks don't give a similar example

 

[praharmitra]

Well, this problem is a direct application of the Residue Theorem.

<br /> \int_C f(z)dz = 2\pi i \sum Res(f)<br />
where
<br /> Res(f) = Lim_{z\rightarrow c}(z-c)f(z)<br />
where c is a simple pole of f. In your case c = 0.

 

[blueyellow]

thanks
but it appeared in a section of book chapters before they even start talking about the residue theorem
so what's the way to do this without using the residue theorem?
the book says i may use the formula:
sin z =sigma from j=0 to infinity [((-1)^j) ( z^(2j+1)))/(2j+1)]!
but how is that supposed to help?
the sigma sign and the j's leave me confused

 

[praharmitra]

sigma stands for summation. Are you aware of the Taylor series expansion of sin(z)? The book wants you to expand, and solve each integral individually

 

[blueyellow]

and from that i should eventually get to the answer 0.how am i supposed to get that? so if i expand, but the series goes on and on, i'll b doing integrals forever...
mayb i don't understand cos i havnt actually tried the question yet. i'll try it.. eventually

 

[praharmitra]

Well, you won't have to do integrals forever.

Try and prove the following -

<br /> \int_C z^n dx = &amp;0,\ n\neq-1; \\ &amp;2\pi i,\ n=-1<br />
That would immediately put all your integrals to zero!