Calculating Complex Integrals using Cauchy Formula on a Circular Path | z = 4

[player1_1_1]

Homework Statement

\oint_{L} \frac{ \mbox{d} z}{ z(z+3) } and L:|z|=4

The Attempt at a Solution

what is assumption, is it oriented positive or negative? and Cauchy formula, can it be done like this?
\frac{ 1 }{ 3 } \left( \oint_{L} \frac{ \mbox{d} z}{ z } - \oint_{L} \frac{ \mbox{d} z}{ z+3 } \right)

 

[HallsofIvy]

If nothing else is said, the integral is assumed to be in the positive orientation, counter-clockwise.
Yes, your partial fraction reduction is correct and the integral can be done in that way. Letting z= e^{i\theta} in the first integral and z= 3+ e^{i\theta} in the second will give very simple integrals, giving the residues at z= 0 and z= 3.

 

[player1_1_1]

thanks for answer, I got 0, is it possible? and please tell me why residue in 3, not -3?

 

[player1_1_1]

up.,

 

[Dick]

thanks for answer, I got 0, is it possible? and please tell me why residue in 3, not -3?

Sure the pole is at -3. And 0 is right. In fact, it's generally true that if f(z) is a polynomial of degree greater than 1 then the integral is zero around a contour that encloses all of the poles.