can somebody explain to me why ∫(1/((z-2)^3)dz=0(integral is bounded by |z-1|=2)?????? z is complex number consder

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- Nov 30th 2012, 08:44 AMcummings123321complex integral
can somebody explain to me why ∫(1/((z-2)^3)dz=0(integral is bounded by |z-1|=2)?????? z is complex number consder

- Nov 30th 2012, 10:45 AMHallsofIvyRe: complex integral
You say the integral is "

**bounded**by |z- 1|= 2" but this is a path integral. Do you mean that |z- 1|= 2**is**the path?

If so, the only point at which is not analytic is at z= 2 which is inside the curve. That means that the integral around any simple closed path that**also**encloses z= 2 will be the same. In particular, the integral around, say |z- 2|= 1, the circle of radius 1 with center at 2, must have the same integral. But on the path, we can use the parameterization with t going from 0 to . Now, , so the integral becomes

Can you see that that is 0?

(More generally, if f(z) has a "pole" at then the integral around will be times the**residue**of f at . However, the "residue" of function f at is the coefficient of the in the Laurent series for f around . Here the entire Laurent series for f around the z= 2**is**that single term . There is NO term. Its coefficient is 0 so the residue is 0 so the integral is 0.) - Nov 30th 2012, 11:18 AMcummings123321Re: complex integral
OK,If i used the way to solve out the case ∫(1/(z-2))dz the path is |z|=1,then let z=exp(it) dz=iexp(it)dt so we get ∫(1/(z-2))dz = ∫(iexp(it)/(exp(it)-2) dt=ln|exp(it)-2|=0 (the integral from 0 to 2pi) is my argument right? i am not sure the step from ∫(iexp(it)/(exp(it)-2) dt to ln|exp(it)-2|here