Real-Variable Expressions of cosh(z) & Finding Complex Zeroes

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Homework Help Overview

The discussion revolves around expressing the hyperbolic cosine function, cosh(z), in terms of real variables and finding its complex zeroes. The problem involves understanding the behavior of cosh(z) when z is a complex number, specifically in the context of complex analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the expression of cosh(z) using real variables, with attempts to derive the function's form and identify its zeroes. Questions arise about the correctness of initial expressions and the implications of certain transformations. There is also discussion on how to represent the zeroes on an Argand diagram.

Discussion Status

The discussion is active, with participants providing insights into the formulation of cosh(z) and its zeroes. Some participants have offered corrections and clarifications regarding the expressions used, while others are seeking further assistance on visual representation in the Argand plane.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance they can provide to one another. There is also a focus on ensuring that the derived expressions and interpretations are mathematically sound.

fredrick08
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Homework Statement


express cosh(z) in terms of functions with real variable only, hence find all complex zeroes and mark on argand diagram.


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 please? 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
 
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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.
 
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?
 
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?
 
please can anyone help?
 

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