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Complex cosh

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    express cosh(z) in terms of functions with real variable only, hence find all complex zeroes and mark on argand diagram.

    3. The attempt at a solution
    I did cosh(z)=cosh(x+iy) then got down to cosh(z)=cos(x)cos(y)-sin(x)sin(y)... can anyone confirm this plz? also i got told, that cosh(z)=0, when z=pi*k*i+pi*i/2....when k is a interger... I have no idea, how my friend got this, of if its even correct, or what i have done is correct either..... but i would appreciate any help, especially with drawing the zeroes on the argand diagram, becasue with the answer i have, there is only 2 per 2pi
  2. jcsd
  3. Mar 14, 2010 #2


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    Well, what you have can't be right because cosh(z) is NOT a real number for all complex z, but cos(x)cos(y)-sin(x)sin(y) is real for all real x and y.

    [tex]cosh(x+iy)= \frac{e^{x+iy}+ e^{-x-iy}}{2}= \frac{e^xe^{iy}+ e^{-x}e^{-iy}}{2}[/tex]
    [tex]= \frac{e^xe^{iy}+ e^xe^{-iy}- e^xe^{-iy}+ e^{-x}e^{-iy}}{2}[/tex]
    [tex]= e^x\frac{e^{iy}+ e^{-iy}}{2}- e^{iy}\frac{e^x+ e^{-x}}{2}[/tex]
    [tex]= e^x cos(y)- cosh(x)e^{-iy}[/tex]

    Of course, [itex]e^{ix}= cos(x)+ i sin(x)[/itex] so [itex]e^{-iy}= cos(y)- i sin(y)[/itex].

    That is, cosh(z)= e^xcos(y)- cosh(x)(cos(y)- i sin(y))= [e^x- cosh(x)]cos(y)- cosh(x)cos(y)]- i cosh(x)sin(y).

    since [itex]cosh(x)= (e^x+ e^{-x})/2[/itex],
    [tex]e^x- cosh(x)= e^x- (e^x+e^{-x})/2= (e^x- e^{-x})/2= sinh(x)[/tex]

    [tex]cosh(z)= sinh(x)cos(y)- i cosh(x)sin(y)[/tex]

    That will be 0 when sinh(x)cos(y)= 0 and cosh(x)sin(y)= 0.
  4. Mar 14, 2010 #3
    ahh i see what i did wrong... i forgot about the ix, and converted the wrong terms... ok.. so buy solving those 2 equations for zero, give x=0 and y=(2k+1)pi/2 and second eq gives, y=k*pi, and x=(2k+1)i*pi/2 where k is a integer. is that right... but how do i put that on an argand diagram?
  5. Mar 14, 2010 #4
    so cosh(z)=0, when z=(2k+1)i*pi/2.... coz that would mean cos((2k+1)pi/2)... which give zero for all half pi's? so is my diagram just a point at iy=I*pi.2 and -I*pi/2.. for the principal values? or is there special way of drawing it?
  6. Mar 14, 2010 #5
    please can anyone help?
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