Complex Analysis

Math 312 Spring 1998  MWF 10:30am - 11:25am
Buckmire  Fowler 112

Class 22 (Monday March 9)

SUMMARY Path Independence and the Cauchy-Goursat Theorem CURRENT READING Brown & Curchill pages 104-111 NEXT READING Brown & Curchill pages 116-123

We have been practicing our complex integration skills. Today we will learn how to use antiderivatives and the Cauchy-Goursat Theorem to evaluate contour integrals more efficiently (simply!) Recall that the value of ∫_{C} 2\overline{z}^{2} dz did depend on the contour when we evaluated it in the previous class. However the value of ∫_{C} 2z^{2} dz did not depend on the contour chosen and in each case the

integral is equal to \frac{−32}{3} How is f(z) = 2z^{2} a different function than g(z) = 2\overline{z}^{2}?

2z^{2} is a  while 2\overline{z}^{2} is a  .

Independence of Path THEOREM: Suppose that the function f(z) is continuous in a domain (open connected set) D and has an antiderivative F (z) throughout D, i.e. dF/dz = f(z) at each point in D. Then for every contour Γ lying in D connecting z_{1} to z_{2} we have the result

∫_{Γ} f(z) dz = F (z_{2}) − F (z_{1})

In otherwords, the value of the integral is independent of the path chosen to link z_{1} and z_{2} So, getting back to the difference between f(z) = 2z^{2} and g(z) = 2\overline{z}^{2} we know that 2z^{2} has the property that it is continuous and is equal to the derivative of 2z^{3}/3 on the open set z ∈ C. g(z) has no similar antiderivative. (This is not surprising, since g(z) doesn’t have a derivative either.)

\underline{Example}

∫_{C} 2z^{2} dz = where C is a contour from 2 to −2 Note that according to the above theorem the function F (z) will be analytic and continuous on the domain D (since it has a derivative at every point of the open set D). A corollary of the above theorem is the famous Cauchy-Goursat Theorem: If f(z) is analytic at all points interior to and on any simple closed contour Γ, then

∮_{Γ} f(z)dz = 0.

(Is this result surprising?) \underline{Groupwork}

1. ∫_{C} sin(iz) + 2e^{z} dz, where Cis a contour joining 0 to πi

2.∮_{C} \frac{dz}{z^{2} − 4} dz, where C : |z| = 1

3. ∫_{−i}^{i} \frac{dz}{z^{2}}  and  ∮_{C_{r}} \frac{dz}{z^{2}} where C_{r} : |z| = 1