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It is often convenient to have a way of bounding the size of an integral, and one such method is theML-inequality.SupposeMis a positive real number such that the functionf(z)satisfies|f(z)| ≤Mfor allzon theimage of a curveCin the plane.If we letLbe the length of the image of the curveC, we have theML-inequality:ZCf≤M·L.Example 4.7.Supposef(z) =3zz2+ 2, andCis the arc of the circle|z|= 2that lies in the first quadrant.We must find an appropriate boundMfor the functionf(z)forzon the arc of the circle.Note that when we estimate the size of a quotient, we look for the maximum of the numerator and theminimum of the denominator.So|f(z)|=3zz2+ 2≤3|z||z|2-2=62= 3 =M,where we have used one form of the triangle inequality: Since|z|= 2, so|z2|=|z|2= 4>2, so|z2+ 2| ≥ |z|2-2 =|z|2-2 = 4-2 = 2on the circle|z|= 2.Then,ZCf≤M·L= 3·2π24= 3π,since the length of one quarter of a circle of radius 2 is equal toπ.Mathematics 366: Complex Variables IStudy Guide89
Example 4.8.SupposeCis the straight line segment from 0 to1 +iandf(z) =ez.Since|ez|=exand the real exponential function is a strictly increasing function, we have|f(z)| ≤e1=e=Mon the image ofC(the maximum value ofexoccurs whenx= 1, its largest value on the image ofC).Thus,ZCf≤M·L=e·√2,where we have found the length ofCby using Pythagoras’ theorem.Incidentally, this integral can be found exactly, since the functionezhas a complex antiderivative, namelyitself!RemarkWe would like to emphasize the fact that the estimate that we obtain from theML-inequality only gives anupper bound for the possible size of the integral, and does not have to be close to its actual value.To illustrate, suppose inExample 4.7, above, we tookCto be the whole circle|z|= 2.Then, its length would be4π, and the sameM= 3would still work.So theML-inequality tells us that the modulus of the integral is less than or equal to12π. However, theactual value of this integral is6πi, so the estimate is not very close! We discuss how to compute this valuelater.We could create more extreme examples, but you should get the point.4.3Contour Integration and Green’s TheoremIn this section, we undertake our first foray into the theory behind Cauchy’s theorem.Indications1.Read Section 4.3, “Contour Integration and Green’s Theorem,” pages 172–179.2.Read the comments below.3.Do Exercises 2–7 and 17–20 on pages 180–181.CommentsFor a closed contour, by default, we always traverse counterclockwise exactly once. For example,|z|= 1,we can parameterize aseiθand letθgoes from 0 to2π, then the corresponding pointz=eiθwill start fromMathematics 366: Complex Variables IStudy Guide90
z= 0, moving along the unit circle counterclockwise for exactly one loop. When a curve is given acounterclockwise direction or can only be traversed counterclockwise, we say it ispositively oriented.