When calculating a complex integral over a closed path does the orientation factor into the calculations?

Question

For example:

let C be a square that is bounded by the lines x= /- 2, y= /- 2 with clockwise orientation. Calculate the closed integral of tan(z)/(z^2).

I was wondering what formula to use, cauchy's or something else?

Answer 1

f(z)=tan(z)/z^2=sin(z)/[cos(z)*z^2]

The loop: x=+-2, y=+-2, CW

f(z) has 3 simple poles inside the closed loop of integration:

z1= -π/2, z2=0 and z3=π/2 ==>

Residue theorem:

` ∫ f(z)*dz= -2πi∑ Res(zi), i=1,2,3 ( - sign because of CW orientation)

(c)

It is positive if the integration is in a counter clockwise

("mathematically positive") manner.

Res(zi)=lim[z-->zi] (z-zi)*f(z), if zi is a simple pole

Res(0)=lim[z-->0] z*f(z)=

lim[x-->0] tan(z)/z=1

Res(-π/2)=lim[z--> -π/2] (z+π/2)*tan(z)/z^2=

lim[z--> -π/2] sin(z)/z^2 * lim[z--> -π/2] (z+π/2)/cos(z)=

-1/(-π/2)^2*lim[z--> -π/2] (z+π/2)'/[cos(z)]'=

(-4/π^2)*lim[z--> -π/2] 1/[-sin(z)]= -4/π^2

Res(π/2)=lim[z-->π/2] (z-π/2)*tan(z)/z^2=

lim[z--> π/2] sin(z)/z^2 * lim[z-->π/2] (z-π/2)/cos(z)=

1/(π/2)^2*lim[z-->π/2] (z-π/2)'/[cos(z)]'=

(4/π^2)*lim[z-->π/2] 1/[-sin(z)]= -4/π^2

` ∫ f(z)*dz= -2πi(1-4/π^2-4/π^2)=i(16/π-2π)

(c)

http://en.wikipedia.org/wiki/Residue_the...