MATH502_HW3 - Serge Ballif(1 Evaluate the integral MATH 502...

Info icon This preview shows pages 1–2. Sign up to view the full content.

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 π 0 cos(cos θ ) cosh(sin θ ) d θ = 2 π. We know that cos( z ) is a holomorphic function on all of . The Cauchy integral formula tells us that 1 · cos( w ) = 1 2 πi H cos( z ) z - w d z , so H cos( z ) z d z = 2 πi . To calculate R 2 π 0 cos(cos θ ) cosh(sin θ ) d θ we will integrate cos( z ) ( z - w ) over the curve γ ( t ) = cos( t ) + i sin( t ) on [0 , 2 π ]. Note that γ 0 ( 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 πi = I cos z z d z = Z 2 π 0 cos(cos t + i sin t ) cos t + i sin t ( - sin t + i cos t ) d t = Z 2 π 0 i cos(cos t + i sin t ) d t = i Z 2 π 0 [cos(cos t ) cos( i sin t ) - sin(cos t ) sin( i sin t )] d t = i Z 2 π 0 cos(cos t ) cosh( - sin t ) d t - i Z 2 π 0 sin(cos t ) sin( i sin t ) d t = i Z 2 π 0 cos(cos t ) cosh(sin t ) d t - 0 . Dividing both sides by i yields the desired result. Note that the calculation R 2 π 0 sin(cos t ) sin( i sin t ) d t
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern