Mathematics of Physics and Engineering

The Theorem and Formula of Cauchy 195

gral over a closed curve C (that is, a curve for which r(a) = r(b)) is often
denoted by §c. The letter z is not the only possible notation for the variable
of integration; in particular, we will often use the Greek letter ( when z is
being used for other purposes.

EXERCISE 4.2.12.C (a) Writing f(z) = u(x,y) + iv(x,y), verify that

/ f(z)dz = / (udx - vdy) + i / (vdx + udy). (4.2.5)

In particular, by (3.1.17) on page 131, change of orientation of the curve
C reverses the sign of fc f(z)dz. (b) Verify that the length of the curve can
be written as fc \dz.

EXERCISE 4.2.13? Let f(z) = (z — z 0 )n, where ZQ is a fixed complex number
and n, an integer: n = 0, ±1,±2,.... Let C be a circle with radius p,
center at z§, and orientation counterclockwise. Verify that §c f(z)dz = 2-ni
if n = — 1 and fc f(z)dz = 0 otherwise. Hint: z(t) = zo + peu, 0 < t < 2TT.

In other words, you show in Exercise 4.2.13 that

Jf

-dz ={—' U h (4.2.6)
z-z 0 \=p (z ~ zo)n [0, n = 0,-1, ±2, ±3,....
Recall that, for a real-valued continuous function h = h(x), we have
| fa h(x)dx\ < (b — a) maxx£[a]f,] |/i(:r)|, and the easiest way to prove this
inequality is to apply the triangle inequality to the approximation of the
integral according to the rectangular rule and then pass to the limit. We
will now derive a similar inequality for the complex integral (4.2.4).
By the left-point rectangular rule, we have

/

J a

b "-1

f(z(t))z{t)dt= lim y]/(z(tfc))i(*fc)(tfc+i-tfc), (4.2.7)

fc=0

where a = to < h < ... < tn — b. Note that |i(£fc)| = ll^(*fc)|| where
r = r(t) is the vector function that defines the curve, and we saw on page
29 that

n-l -fe

lim y)||r(tfc)||(tfc+i-tfe)= / ||r(t)||dt = Lc(a,6),

maxltfc+i-tfcHOj^ Ja

the length of the curve C. We then apply the triangle inequality on the
right-hand side of (4.2.7) and denote by maxzec \f{z)\ the maximal value