Solving Complex Hyperbolic functions

  • #1

Homework Statement


I am a little confused on the steps to take to solve these kinds of functions.

Solve:

cosh z = 2i



The Attempt at a Solution



We were given identities for sinh z = 0 and cosh z = 0 and also other identities like

cosh(z) = cos (iz)

So cos (iz) = 2i

cos (z) would be 2i for (Pi/2 + 2kPi) right? So cos(iz) = i(Pi/2 + 2kPi) = 2i?

However, this doesn't look right.
 

Answers and Replies

  • #2
Office_Shredder
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cos(z)=2i definitely does not happen when z=pi/2+2kpi. cos(pi/2)=0 for example.

How you then proceed from there to write down cos(iz) = i(Pi/2 + 2kPi) is unclear
 
  • #3
392
17

Homework Statement


I am a little confused on the steps to take to solve these kinds of functions.

Solve:

cosh z = 2i



The Attempt at a Solution



We were given identities for sinh z = 0 and cosh z = 0 and also other identities like

cosh(z) = cos (iz)

So cos (iz) = 2i

Okay. Now use an equation to convert cosine into exponentials and solve for z using complex logs.
 
  • #4
Okay. Now use an equation to convert cosine into exponentials and solve for z using complex logs.

By "an equation" do you mean some identity I've missed out somewhere?
 
  • #5
392
17
Yes, an identity. It relates cosine to exponentials, it can be derived from euler's formula, I'm sure you have gone over it before.

##\displaystyle \frac{e^{iz} + e^{-iz}}{2} = \cos{z}##
 
  • #6
Ohh that one! Ok, I will try now.
 
  • #7
Since we know that

[itex]\cos \left( x+iy \right) =1/2\,{{\rm e}^{iz}}+1/2\,{{\rm e}^{-iz}}[/itex]

Splitting up the exponents at the top:

[itex]\cos \left( x+iy \right) =1/2\,{{\rm e}^{-y}}{{\rm e}^{ix}}+1/2\,{
{\rm e}^{y}}{{\rm e}^{-ix}}
[/itex]

By this token, cos (iz) = cos(-y + ix) =

[itex]1/2\,{{\rm e}^{-iy}}{{\rm e}^{-x}}+1/2\,{{\rm e}^{iy}}{{\rm e}^{x}}[/itex]

Which simplifies to:

[itex]1/2\,{{\rm e}^{iz}}+1/2\,{{\rm e}^{-iz}}=2\,i[/itex]

Now I am left with:

[itex]{{\rm e}^{iz}}+{{\rm e}^{-iz}}=4\,i[/itex]

I am not sure entirely sure how to apply the log laws here.

I broke the exponentials up again:

[itex]{{\rm e}^{y}}+{{\rm e}^{-ix}}+{{\rm e}^{-y}}+{{\rm e}^{ix}}=4\,i[/itex]

log (4i) would be ln (4) + i(Pi/2 + 2kPi)
 
  • #8
Any help would be appreciated. Got a test tomorrow, and I'm unfortunately struggling with the trig and hyperbolic side of things.
 
  • #9
392
17
Don't do cos(x + iy). Do cos(iz) since that is what you want. Then multiply each side of the equation by e^z and use the quadratic formula to solve for e^z. Once you have that, take the log of both sides to solve for z!
 
  • #10
I am not sure how to get the cos (iz) identity, it might be where I'm going wrong. I just realised my identity above for cos (iz) is the same as for cos (z) so obviously that can't be right.

cos (iz) = cos (-y + ix) yes?

What I did now was:

[itex]1/2\,{{\rm e}^{-z}}+1/2\,{{\rm e}^{z}}=2\,i[/itex]

Multiplying through by e^z and simplifying:

[itex]{{\rm e}^{2\,z}}-4\,i{{\rm e}^{z}}+1=0[/itex]

Is this more along the lines? I've never actually seen this before in the lectures or in my notes, which is why I'm a bit slow at getting the point.
 
  • #11
If that's correct. Solving for e^z leads to 2i ± i√5

Taking logs means z = ln(4.23....) ± i(Pi/2 + kPi)

The answer given is:

[itex] \left( -1 \right) ^{n}{\it arcsinh} \left( 2 \right) +i \left( n+1/2
\right) \pi
[/itex]

I can see it's the same answer now, checking the value of arcsinh. How did they get that into that form though. I wouldn't have recognised the ln (4.23...) turning to arcsinh(2)

Thanks a lot for your help though, it is really appreciated. I was struggling with this without an example to go off.
 
Last edited:
  • #12
392
17
If that's correct. Solving for e^z leads to 2i ± i√5

Taking logs means z = ln(4.23....) ± i(Pi/2 + kPi)

The answer given is:

[itex] \left( -1 \right) ^{n}{\it arcsinh} \left( 2 \right) +i \left( n+1/2
\right) \pi
[/itex]

I can see it's the same answer now, checking the value of arcsinh. How did they get that into that form though. I wouldn't have recognised the ln (4.23...) turning to arcsinh(2)

Thanks a lot for your help though, it is really appreciated. I was struggling with this without an example to go off.

Sorry for the late reply. What I got for z is ##z = ln|2 \pm \sqrt{5}| + i arg(2i \pm i \sqrt{5}) = ln|2 \pm \sqrt{5}| + i(\frac{\pi}{2} + 2 \pi k)##, where k depends on the branch of the logarithm.

Now the arcsine hyperbolic bit.. Looking online at the definition of arcsinh, ##arcsinh(x) = ln(x + \sqrt{1 + x^2})##. What we have is ##ln(2 \pm \sqrt{5}) = ln|2 \pm \sqrt{1 \pm 2^2}|##. Here is my thinking. After running ##ln|2 \pm \sqrt{5}|## through wolfram alpha, they are opposite in sign to each other, +1.44... and -1.44.... I'm not sure why they can simplify ##ln|2 \pm \sqrt{1 \pm 2^2}| \text{ to } ln|2 + \sqrt{1 \pm 2^2}|## but that is what they did. Someone else will have to explain that.

EDIT: I forgot about the ##(-1)^n## in the answer. That accounts for the negative sign that they drop so they can express the answer in terms of arcsinh.

Honestly, unless your teacher doesn't like plus or minus signs entering in the answer, I would give my answer as: ##ln|2 \pm \sqrt{5}| + i (\frac{\pi}{2} + 2 \pi k)##. This answer is simply derived from taking ##log(2i \pm i \sqrt{5})##
 
  • #13
I just found out today that both are correct, but taking logs can be a bit of a a headache due to the imaginary component the 2kPi jazz.

Apparently, the way the arcsinh was obtained was using the definition of coshz = cosh x cos y + i sinhx siny

And then equating the real and imaginary components and solving it that way. Which I haven't done it yet, but I saw the demonstration and it made sense.
 

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