Complex Equation Solution: Finding Values for Z using Trig Identities

  • Thread starter Phillipe
  • Start date
In summary: I think I just saw it. I get:e^2^x-e^x*2cos(3)-1 = 0which stinks of quadratic equation.Things could get messy with that cos(3), though, but I'll take a shot. Thanks a lot, dude!
  • #1
Phillipe
10
0

Homework Statement


Find all the values of Z which satisfy the equation, Z being a complex number in the form x+i*y.


Homework Equations


Every trig identity out there.


The Attempt at a Solution


Here's what I got so far:

Cosh(z) = i*Cos(3)
Cosh(x)*Cos(y)+i*(Sinh(x)*Sin(y)) = i*Cos(3)

Therefore,

Cosh(x)*Cos(y) = 0
Sinh(x)*Sin(y) = i*Cos(3)

y = Pi/2 (since there's no value of x which can give Cosh(x) = 0). y may also be Pi/2 + 2nPi, where n is an integer.

Therefore,
Sinh(x) = i*Cos(3)

But I don't know how to proceed from there.
 
Physics news on Phys.org
  • #2
Use the identity sinh(x) = .5[exp(x)-exp(-x)]. Identities for sinh(x), and cosh(x) can be found on wikipedia.
 
  • #3
Samuelb88 said:
Use the identity sinh(x) = .5[exp(x)-exp(-x)]. Identities for sinh(x), and cosh(x) can be found on wikipedia.

I have thought of using that, but I don't know how to get the values for x from there. Would you be so kind to tell me how?
 
  • #4
Phillipe said:
I have thought of using that, but I don't know how to get the values for x from there. Would you be so kind to tell me how?

Sure. After we replace sinh(x) with .5[exp(x)-exp(-x)], let's first multiply both sides of the equation by 2, then by e^x. Set everything equal to zero (isolate all terms on one side) which should yield:
[tex]e^2^x-something*e^x-1=0[/tex]
Can you see what to do from here?
 
  • #5
Samuelb88 said:
Sure. After we replace sinh(x) with .5[exp(x)-exp(-x)], let's first multiply both sides of the equation by 2, then by e^x. Set everything equal to zero (isolate all terms on one side) which should yield:
[tex]e^2^x-something*e^x-1=0[/tex]
Can you see what to do from here?

I think I just saw it. I get:

[tex]e^2^x-e^x*2cos(3)-1 = 0[/tex]

which stinks of quadratic equation.Things could get messy with that cos(3), though, but I'll take a shot. Thanks a lot, dude!
 

What is an equation?

An equation is a mathematical statement that shows the equality between two quantities or expressions. It typically includes variables, numbers, and mathematical operations.

How do I solve an equation?

To solve an equation, you need to isolate the variable on one side of the equation by using algebraic operations such as addition, subtraction, multiplication, or division. The goal is to get the variable by itself on one side of the equation, with all the constants on the other side.

What is the order of operations in solving an equation?

The order of operations in solving an equation is the same as the order of operations in mathematics, which is PEMDAS (parentheses, exponents, multiplication and division from left to right, and addition and subtraction from left to right).

What if I can't solve an equation?

If you are having trouble solving an equation, try breaking it down into smaller steps or using different algebraic techniques such as factoring, substitution, or solving for one variable at a time. If you are still having difficulty, consult a math tutor or teacher for assistance.

Why is it important to solve equations?

Solving equations is important in many fields, including science, engineering, and economics. It allows us to find unknown quantities, make predictions, and solve real-world problems. It also helps us develop critical thinking and problem-solving skills.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
370
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
693
  • Calculus and Beyond Homework Help
Replies
6
Views
663
  • Calculus and Beyond Homework Help
Replies
6
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
474
  • Calculus and Beyond Homework Help
Replies
6
Views
677
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
27
Views
542
Back
Top