Complex Analysis - Solving Complex Trig functions

In summary, the homework statement is about two methods for solving a homework problem involving logarithms and exponential equations. One method is simpler and easier to understand, while the other is more complex but may lead to a more accurate solution.
  • #1
NewtonianAlch
453
0

Homework Statement


Now, I know there's two ways to go about this and it seems everywhere I look around on the web people are solving it in a way I think that seems longer, harder and more prone to mistakes in exams. It involves using the exponential identities and taking logs. I was shown another way, but unfortunately I haven't quite got the grasp of it.

sinh z = 0

sinh (x + iy) = sinh(x)cos(y) + i cosh(x)sin(y) = 0

So

1) sinh(x)cos(y) = 0
2) cosh(x)sin(y) = 0

(1) Either x = 0 or y = ± Pi/2
(2) cosh (x) is never 0, so therefore x is not 0. Hence y = 0

This is where I'm stuck, I do not know how to go from here.

The answer is z = i*k*Pi
 
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  • #2
You have two simultaneous equations. You've correctly deduced that cosh(x) is never 0, so therefore in equation 2 divide both sides by cosh(x) to get

sin(y) = 0

which occurs when y = k*Pi, where k is an element of the integer set

Ok so we now know y! Let's look at x. Now you use your first equation:

at y = k*Pi, cos(y) = 1 or -1

so divide both sides by cos(y) and get

sinh(x) = 0, therefore x = 0

so what we have is z = x + iy = i * k * Pi.

This method is a lot more simple, it just involves working through the simultaneous equations carefully. Where you started to go wrong was when you assumed y just has to equal 0, rather than the general solution of k * Pi
 
  • #3
Hmm, interesting. I think I'm starting to see it now.

So suppose I have cosh z = 1

cosh (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 1

cosh(x)cos(y) = 1
sinh(x)sin(y) = 0

cosh(x) = 1 only when x = 0, and cos (y) = 1 for 2*k*Pi

So now we already have values for x and y, substituting this into the imaginary component, validates that.

So z = i*2*k*Pi

Is my reasoning correct? Thank you for your help by the way.
 
  • #4
Stuck again now!

cosh z = 2i

I posted this question earlier but back then I only knew the exponential and log method.

What I've got so far:

cos (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 2i

1) cosh(x)cos(y) = 0
2) sinh(x)sin(y) = 2i

From equation 1, cosh x cannot be zero. For cos (y) = 0, y is periodic so y = (Pi/2 + kPi)

Now this is where I'm stuck on how to deal with that in equation 2.

The final answer given is (-1)^k arcsinh(2) + i(kPi +Pi/2)

I can see how that answer for y I got in equation 1 has something to do with the final answer.

Edit:

I'm thinking that if y = (Pi/2 + kPi), this would mean sin(y) = (-1)^k

So dividing sin (y) in equation 2, we get sinh(x) = (-1)k 2i

Taking arcsinh of this now means arcsinh((-1)^k 2i)

Does this mean I can take out the (-1)^k ? I still haven't figured out what to do with the i hanging around.
 
Last edited:
  • #5
NewtonianAlch said:
cosh(x) = 1 only when x = 0, and cos (y) = 1 for 2*k*Pi

That's true but cosh(x) cos(y) = 1 holds for other values as well, for example cosh(x)=2, cos(y) = 1/2.

2) sinh(x)sin(y) = 2i

that should be 2, not 2i
 
  • #6
clamtrox said:
That's true but cosh(x) cos(y) = 1 holds for other values as well, for example cosh(x)=2, cos(y) = 1/2.



that should be 2, not 2i

That would explain it then, thanks!
 

Related to Complex Analysis - Solving Complex Trig functions

What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of complex numbers and their properties. It involves the use of calculus and algebra to analyze functions that are defined on the complex plane.

What are complex trigonometric functions?

Complex trigonometric functions are functions that involve complex numbers and trigonometric functions such as sine, cosine, tangent, etc. These functions are defined on the complex plane and have both a real and imaginary component.

How do you solve complex trigonometric functions?

To solve complex trigonometric functions, you can use the properties of complex numbers and trigonometric identities. You can also use techniques such as converting complex numbers to polar form, using Euler's formula, and applying the laws of exponents.

What is the importance of solving complex trigonometric functions?

Solving complex trigonometric functions is important in many areas of mathematics and science, such as engineering, physics, and signal processing. It allows for a deeper understanding of the behavior of complex systems and can be used to solve practical problems.

Are there any applications of complex trigonometric functions in real life?

Yes, complex trigonometric functions have many applications in real-life situations. For example, they are used in electrical engineering to analyze alternating current circuits, in physics to study wave motion, and in navigation to calculate the position of objects using complex coordinates.

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