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

In summary, the student is trying to find all complex zeroes on the Argand diagram for cosh(z). They are also trying to find when cosh(z)=0. They are having trouble with the math and need help with the zeroes and drawing the diagram.
  • #1
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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  • #2
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.
 
  • #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?
 
  • #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?
 
  • #5
please can anyone help?
 

1. What is cosh(z)?

Cosh(z) is a hyperbolic function that is defined as the ratio of the adjacent side to the hypotenuse of a right triangle with a hyperbolic angle z. It is also known as the hyperbolic cosine function.

2. How is cosh(z) related to complex numbers?

Cosh(z) can be expressed as a complex variable expression by using Euler's formula, which states that e^(iz) = cos(z) + i*sin(z). Therefore, cosh(z) can be written as (e^z + e^(-z))/2, where z is a complex number.

3. How do you find the complex zeroes of cosh(z)?

The complex zeroes of cosh(z) can be found by setting the expression equal to zero and solving for z. This can be done by using the quadratic formula or by factoring the expression.

4. Can cosh(z) have imaginary zeroes?

Yes, cosh(z) can have imaginary zeroes, as it is a complex-valued function. This means that the zeroes can have both real and imaginary parts.

5. What are the applications of cosh(z) in real-world problems?

Cosh(z) has various applications in physics, engineering, and mathematical modeling. It is used to describe the shape of a hanging cable, the behavior of a spring, and the motion of a pendulum. It is also used in solving differential equations and in statistical mechanics.

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