11.7. EXERCISES 273

complex, explain why sin(z) cannot be bounded. Hint: Use the theorem about thezero set having a limit point on a connected open set.

14. The functions z→ sin(z2),z→ cos

(z2), and z→ eiz2

are all analytic functions sincethe chain rule continues to hold for functions of a complex variable. This problem ison the Fresnel integrals using contour integrals. In this case, there is no singular partof the function. The contour to use is

x

y

Then using this contour and the integral∫

0 e−t2dt =

√π

2 , explain why

0 =∫

γr

eiz2dz+

∫ r

0eix2

dx−∫ r

0ei(

t(

1+i√2

))2(1+ i√2

)dt

=∫

γr

eiz2dz+

∫ r

0eix2

dx−∫ r

0e−t2

(1+ i√

2

)dt

=∫

γr

eiz2dz+

∫ r

0eix2

dx−√

π

2

(1+ i√

2

)+ e(r)

Where limr→0 e(r) = 0. Now examine the first integral. Explain the following stepsand why this integral converges to 0 as r→ ∞.∣∣∣∣∫

γr

eiz2dz∣∣∣∣= ∣∣∣∣∫ π

4

0ei(reit)

2rieitdt

∣∣∣∣≤ r∫ π

4

0e−r2 sin2tdt =

r2

∫ 1

0

e−r2u√

1−u2du

=r2

∫ r−(3/2)

0

1√1−u2

du+r2

(∫ 1

0

1√1−u2

)e−(r1/2)

15. If γ : C→ C is a parametrization of a curve with γ being differentiable, one to oneon (a,b) ,a < b with continuous derivative, the length of C is defined as

sup

{∑P|γ (ti)− γ (ti−1)| ,P a partition of [a,b]

}Show this is independent of equivalent smooth parametrization and that in every case,it equals

∫ ba |γ ′ (t)|dt, the integral of the absolute value of the derivative.

16. Consider the following contour consisting of the orientation shown by the arrows.

γr

γR

There is a large semicircle on the top of radius R and a small one of radius r. If γ refersto the piecewise smooth, oriented contour consisting of the two straight lines and two

11.7. EXERCISES 27314.15.16.complex, explain why sin(z) cannot be bounded. Hint: Use the theorem about thezero set having a limit point on a connected open set.The functions z > sin (2?) ,Z—> COS (2?) , and z—> e are all analytic functions sincethe chain rule continues to hold for functions of a complex variable. This problem ison the Fresnel integrals using contour integrals. In this case, there is no singular partof the function. The contour to use isaN xThen using this contour and the integral {9° e~‘ "dt = ve explain whyr ri. i 2 ;[ eaz+ | eax | (tS) (=) dtriz2 " ix? r 1? 1+id + d. - | ( dtJe Zz 0 e xX P e V2+2 ro.o Ji 1+ied +f e* ax (Et) te r[ act | 2 (>) teWhere lim,_,9 e (r) = 0. Now examine the first integral. Explain the following stepsand why this integral converges to 0 as r > ©,a 4a) "Af it\2 . a 2:| e dz| = i eilre") rie dt < rf er sin 2t yy _v, 0 0rolIr pl ererp ery2/0 V1—u2r r— (3/2)r 1 12 Jo V1-u2 " 5(/ ana)If y: C > C is a parametrization of a curve with y being differentiable, one to oneon (a,b) ,a < b with continuous derivative, the length of C is defined as-(r'2)sup s ly(t;) — y(t-1)|,P a partition of [a,b]PShow this is independent of equivalent smooth parametrization and that in every case,it equals f. iM |y (t)|dt, the integral of the absolute value of the derivative.Consider the following contour consisting of the orientation shown by the arrows.YReLyThere is a large semicircle on the top of radius R and a small one of radius r. If y refersto the piecewise smooth, oriented contour consisting of the two straight lines and two