Cauchy Integral with absolute value in simple curve

[Bacat]

Homework Statement

Integrate using Cauchy Formula or Cauchy Theorem:

I=\oint_{|z+1|=2}\frac{z^2}{4-z^2} dz

(From Complex Variables, Stephen Fisher (2nd Edition), Exercise 2.3.3)

Homework Equations

\frac{1}{2\pi i}\oint_{\gamma} \frac{f(s)}{s-z} ds= \left\{ \begin{array}{lr}f(z) & : z \in \gamma\\0 & : z \notin \gamma \end{array} where \gamma is defined as the interior of the simple closed curve described by the line integral.

The Attempt at a Solution

I = \oint_{|z+1| = 2} \frac{\left(\frac{z^2}{z-2}\right)}{z+2}} dz = 2\pi i (f(z)) = 2 \pi i

Correct solution is 2\pi i

Question

I understand that f(z)=1 for this problem, but I don't know why. I tried drawing a graph of the function but I'm confused about the |z+1|=2 part. This says that z=1 or z=-3, but this doesn't seem to be a simple closed curve anymore.

I have singularities at z=\pm2 but I'm not sure which singularity is included in the line integral because I don't know how to draw the graph.

Can someone help?

 

[Vagrant]

I'm confused about the |z+1|=2 part. This says that z=1 or z=-3, but this doesn't seem to be a simple closed curve anymore.

|z+1|=2, means that it is a circle with center at (-1,0) and radius = 2 in the Argand plane.

 

[HallsofIvy]

Homework Statement

Integrate using Cauchy Formula or Cauchy Theorem:

I=\oint_{|z+1|=2}\frac{z^2}{4-z^2} dz

(From Complex Variables, Stephen Fisher (2nd Edition), Exercise 2.3.3)

Homework Equations

\frac{1}{2\pi i}\oint_{\gamma} \frac{f(s)}{s-z} ds= \left\{ \begin{array}{lr}f(z) & : z \in \gamma\\0 & : z \notin \gamma \end{array} where \gamma is defined as the interior of the simple closed curve described by the line integral.

The Attempt at a Solution

I = \oint_{|z+1| = 2} \frac{\left(\frac{z^2}{z-2}\right)}{z+2}} dz = 2\pi i (f(z)) = 2 \pi i

Correct solution is 2\pi i

Question

I understand that f(z)=1 for this problem, but I don't know why. I tried drawing a graph of the function but I'm confused about the |z+1|=2 part. This says that z=1 or z=-3, but this doesn't seem to be a simple closed curve anymore.

On the real line, |z+1|= 2 is true for z= 1 and z= -3 but the complex plane is a plane including points both above and below the real line. Geometrically |z- a| can be interpreted as distance from z to a in the complex plane and so the set of points, z, satisfying |z-a|= r is the set of points that have distance r from a: a circle with radius r and center a. The set of points, z, satisfying |z+1|= |z- (-1)|= 2 is the circle with center at -1 and radius 2. The real line is a diameter of that circle and z= 1 and z= -3 are the endpoints of that diameter.

I have singularities at z=\pm2 but I'm not sure which singularity is included in the line integral because I don't know how to draw the graph.

Can someone help?

If z= 2, then |z+ 1|= |2+ 1|= 3 which is outside the circle. If z= -2, then |z+ 1|= |-2+1|= 1 which in inside the circle.

 

[Bacat]

Thank you! I understand now.