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Math Help - complex function

  1. #1
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    complex function

    Hi everybody,

    F(z)= e^x(cosy+isiny), i must show that F( z_1+z_2)=F( z_1).F( z_2)
    can you help me please???
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  2. #2
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    You'll find it a lot easier if you recognize that \cos{y}+i\sin{y}=e^{iy}.

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

    - Hollywood
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  3. #3
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    The hard way to do it, if you don't have that formula yet, is to use trig identities.

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

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

    Also, cos(y_1+ y_2)= cos(y_1)cos(y_2)- sin(y_1)sin(y_2) and sin(y_1+ y_2)= sin(y_1)cos(y_2)+ cos(y_1)sin(y_2) so that cos(y_1+ y_2)+ i sin(y_1+ y_2) = 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) = (cos(y_1)+ i sin(y_1))(cos(y_2)+ i sin(y_2))
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  4. #4
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    Quote Originally Posted by bhitroofen01 View Post
    Hi everybody,

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


    Well, write 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 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 \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]
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