Results 1 to 3 of 3

Math Help - Trig proof -- Complex analysis

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    51

    Trig proof -- Complex analysis

    Prove that |cos^2 (z)|+|sin^2 (z)| = 1 is false if z=x+iy with y\ne 0.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by JJMC89 View Post
    Prove that |cos^2 (z)|+|sin^2 (z)| = 1 is false if z=x+iy with y\ne 0.

    Write \cos z=\frac{e^{iz}+e^{-iz}}{2}\,,\,\,\sin z=\frac{e^{iz}-e^{-iz}}{2i} , so

    =\cos^2z=\frac{e^{2iz}+e^{-2iz}+2}{4}=\frac{1}{4}\left[e^{-2y}\left(\cos 2x+i\sin 2x\right)+e^{2y}\left(\cos 2x-i\sin 2x\right)\right]+\frac{1}{2}=

    = \frac{1}{2}\left[\cos 2x\cosh 2y+1+i\sin 2x\sinh 2y\right]

    \sin^2z=\frac{e^{2iz}+e^{-2iz}-2}{-4}=-\frac{1}{4}\left[e^{-2y}\left(\cos 2x+i\sin 2x\right)+e^{2y}\left(\cos 2x-i\sin 2x\right)\right]+\frac{1}{2}=

    =-\frac{1}{2}\left[\cos 2x\cosh 2y-1+i\sin 2x\sinh 2y\right] , so

    |\cos^2z|+|\sin^2z|=\frac{1}{4}\left[(\cos 2x\cosh 2y+1)^2+(\cos 2x\cosh 2y-1)^2+2\sin^22x\sinh^22y\right]

    take it from here (and check the above carefully for mistakes)

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    51
    Thanks

    I got it now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  2. Complex Analysis Proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 1st 2011, 03:56 AM
  3. Complex Analysis: Trig Functions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 29th 2010, 09:09 AM
  4. Complex Analysis: hyperbolic trig and exponential
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 19th 2010, 05:15 PM
  5. Complex Analysis proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: December 14th 2009, 05:19 PM

Search Tags


/mathhelpforum @mathhelpforum