Results 1 to 4 of 4

Thread: complex function

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    40

    complex function

    Hi everybody,

    F(z)=$\displaystyle e^x$(cosy+isiny), i must show that F($\displaystyle z_1+z_2$)=F($\displaystyle z_1$).F($\displaystyle z_2$)
    can you help me please???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    Posts
    1,036
    Thanks
    272
    You'll find it a lot easier if you recognize that $\displaystyle \cos{y}+i\sin{y}=e^{iy}$.

    Post again if you can't finish it from there.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,793
    Thanks
    3035
    The hard way to do it, if you don't have that formula yet, is to use trig identities.

    $\displaystyle F(z_1+ z_2)= e^{x_1+ x_2}(cos(y_1+ y2)+ i sin(y_1+ y_2))$

    Of course, $\displaystyle e^{x_1+ x_2}= e^{x_1}e^{x_2}$.

    Also, $\displaystyle cos(y_1+ y_2)= cos(y_1)cos(y_2)- sin(y_1)sin(y_2)$ and $\displaystyle sin(y_1+ y_2)= sin(y_1)cos(y_2)+ cos(y_1)sin(y_2)$ so that $\displaystyle cos(y_1+ y_2)+ i sin(y_1+ y_2)$$\displaystyle = cos(y_1)cos(y_2)- sin(y_1)sin(y_2)+ i sin(y_1)cos(y_2)+ i cos(y_1)sin(y_2)$$\displaystyle = (cos(y_1)+ i sin(y_1))(cos(y_2)+ i sin(y_2))$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by bhitroofen01 View Post
    Hi everybody,

    F(z)=$\displaystyle e^x$(cosy+isiny), i must show that F($\displaystyle z_1+z_2$)=F($\displaystyle z_1$).F($\displaystyle z_2$)
    can you help me please???


    Well, write $\displaystyle z_1=x_1+iy_1\,,\,z_2=x_2+iy_2\,,\,\,x_1,x_2,y_1,y_ 2\in\mathbb{R}$ , and now just evaluate $\displaystyle F(z_1+z_2)=F((x_1+x_2)+i(y_1+y_2))=e^{x_1+x_2}\lef t[\cos(y_1+y_2)+i\sin(y_1+y_2)\right]$ .

    Now just a little algebra + some powers arithmetic + basic trigonometry to get that the above certainly equals $\displaystyle \left[e^{x_1}(\cos y_1 +i\sin y_1)\right]\cdot \left[e^{x_2}(\cos y_2+i\sin y_2)\right]$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 05:30 AM
  2. [SOLVED] complex analysis: differentiability of a complex function?
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: Feb 16th 2011, 08:33 AM
  3. Complex function help
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Sep 17th 2010, 08:59 AM
  4. Complex function
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Jun 22nd 2010, 04:23 AM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum