Exercise 6.18

Exercise 18: Let γ 1 , γ 2 , γ 3 be curves in the complex plane, defined on [ 0 , 2 π ] by

γ 1 ( t ) = e it , γ 2 ( t ) = e 2 it , γ 3 ( t ) = e 2 πit sin ( 1 t ) .

Show that these three curves have the same range, that γ 1 and γ 2 are rectifiable, that the length of γ 1 is 2 π , that the length of γ 2 is 4 π , and that γ 3 is not rectifiable.

Answers

Since for any x R we have | e ix | = | cos x + i sin x | = cos 2 x + sin 2 x = 1 , the images of all three curves lie on the unit circle S 1 . And if z = cos x + i sin x is an arbitrary point of S 1 , 0 x < 2 π , then z = γ 1 ( x ) = γ 2 ( x 2 ) , so the images of γ 1 and γ 2 are all of S 1 . Also, since

2 π ( π ) sin ( 1 π ) 6.18 2 π ( 2 3 π ) sin ( 3 π 2 ) 1.33 ,

by the intermediate-value theorem (Theorem 4.23) if 1.33 y 6.18 , there is a 0 t 2 π such that 2 πt sin ( 1 t ) = y , so that γ 3 ( t ) = e iy = z . Hence the image of γ 3 is also all of S 1 .

Since | γ 1 ( t ) | = | i e it | = 1 and | γ 2 ( t ) | = | 2 i e 2 it | = 2 , by Theorem 6.27 we have

Λ ( γ 1 ) = 0 2 π 1 dt = 2 π Λ ( γ 2 ) = 0 2 π 2 dt = 4 π .

For γ 3 we have

γ 3 ( t ) = 2 πi ( sin ( t 1 ) t 1 cos ( t 1 ) ) γ 3 ( t ) | γ 3 ( t ) | = 2 π | sin ( t 1 ) t 1 cos ( t 1 ) |

Let a n = ( 2 n + 1 ) π , b n = ( 2 n + 1 2 ) π for any integer n 1 . Then [ a n 1 , b n 1 ] is a subinterval of [ 0 , 2 π ] on which sin ( 1 t ) 0 and cos ( 1 t ) 0 . Hence the length of γ 3 on this subinterval is

a n 1 b n 1 | γ 3 ( t ) | dt = 2 π a n 1 b n 1 sin ( t 1 ) t 1 cos ( t 1 ) dt = 2 π a n b n u 2 sin ( u ) + u 1 cos ( u ) du = 2 π ( b n 1 sin ( b n ) a n 1 sin ( a n ) ) = 1 n + ( 1 4 )

Since Λ ( γ 3 ) must be larger than the divergent sum of such terms, γ 3 is not rectifiable.

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2023-08-07 00:00
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