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Unformatted text preview: Serge Ballif MATH 502 Homework 3 February 8, 2008 (1) Evaluate the integral I cos z z d z around the unit circle. Deduce that Z 2 cos(cos )cosh(sin )d = 2 . We know that cos( z ) is a holomorphic function on all of C . The Cauchy integral formula tells us that 1 cos( w ) = 1 2 i H cos( z ) z w d z , so H cos( z ) z d z = 2 i . To calculate R 2 cos(cos ) cosh(sin ) d we will integrate cos( z ) ( z w ) over the curve ( t ) = cos( t ) + i sin( t ) on [0 , 2 ]. Note that ( t ) = sin( t ) + i cos( t ). We will also use the three trigonometric identities cos( x + y ) = cos x cos y sin x sin y , cos( z ) = cosh( iz ), and cosh( z ) = cosh( z ). 2 i = I cos z z d z = Z 2 cos(cos t + i sin t ) cos t + i sin t ( sin t + i cos t ) d t = Z 2 i cos(cos t + i sin t ) d t = i Z 2 [cos(cos t ) cos( i sin t ) sin(cos t ) sin( i sin t )] d t = i Z 2 cos(cos t ) cosh( sin t ) d t i Z 2 sin(cos t ) sin( i sin t ) d t = i Z 2...
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This note was uploaded on 04/01/2008 for the course MATH 502 taught by Professor Johnridener during the Summer '08 term at Pennsylvania State University, University Park.
 Summer '08
 JOHNRIDENER
 Math, Unit Circle

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